Chapter 27: Atoms

Plum Pudding Model

Model Description

• Proposed by J.J. Thomson in 1898.
• Atom consists of a sphere of positive charge with electrons embedded like seeds in a watermelon.
• Positive charge is uniformly distributed throughout the atom's volume.
• Electrons are negatively charged and scattered inside the positive sphere.
• Model was later disproved by experiments showing different charge distribution.

Solved Examples

• Ex 1: Identify the charge distribution in the plum pudding model.
  Answer: Positive charge is spread uniformly; electrons embedded inside.

Practice Set

• Q1: What does the plum pudding model propose about the atom?
  A. Nucleus at center
  B. Electrons orbit nucleus
  C. Positive charge spread uniformly with embedded electrons
  D. Electrons outside nucleus
  
• Q2: According to the plum pudding model, electrons are:
  A. Orbiting nucleus
  B. Embedded in positive charge
  C. Free particles
  D. Absent
  
• Q3: Which experiment disproved the plum pudding model?
  A. Photoelectric effect
  B. Rutherford's alpha scattering
  C. Compton scattering
  D. Davisson-Germer experiment
  

Answer Key

• Q1: C - Positive charge is uniformly spread with electrons embedded inside.
• Q2: B - Electrons are embedded in the positive charge sphere.
• Q3: B - Rutherford's alpha scattering experiment disproved this model.

Memory Aids

• Think "plums in pudding" for electrons in positive charge.
• Positive charge = pudding; electrons = plums.

Alpha Particle Scattering

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from ²¹⁴₈₃Bi source directed at thin gold foil (thickness 2.1 × 10^-7 m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles N ∝ 1/sin⁴(θ/2).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ 1/8000 of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles N varies with scattering angle θ as:
  A. N ∝ sin² θ
  B. N ∝ cos² θ
  C. N ∝ 1/sin⁴(θ/2)
  D. N ∝ tan θ
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as 1/sin⁴(θ/2).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• N ∝ 1/sin⁴(θ/2) curve shape.

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from ²¹⁴₈₃Bi source directed at thin gold foil (thickness 2.1 × 10^-7 m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles N ∝ 1/sin⁴(θ/2).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ 1/8000 of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles N varies with scattering angle θ as:
  A. N ∝ sin² θ
  B. N ∝ cos² θ
  C. N ∝ 1/sin⁴(θ/2)
  D. N ∝ tan θ
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as 1/sin⁴(θ/2).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• N ∝ 1/sin⁴(θ/2) curve shape.

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from ²¹⁴₈₃Bi source directed at thin gold foil (thickness 2.1 × 10^-7 m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles N ∝ 1/sin⁴(θ/2).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ 1/8000 of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles N varies with scattering angle θ as:
  A. N ∝ sin² θ
  B. N ∝ cos² θ
  C. N ∝ 1/sin⁴(θ/2)
  D. N ∝ tan θ
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as 1/sin⁴(θ/2).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• N ∝ 1/sin⁴(θ/2) curve shape.

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from ²¹⁴₈₃Bi source directed at thin gold foil (thickness 2.1 × 10^-7 m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles N ∝ 1/sin⁴(θ/2).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ 1/8000 of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles N varies with scattering angle θ as:
  A. N ∝ sin² θ
  B. N ∝ cos² θ
  C. N ∝ 1/sin⁴(θ/2)
  D. N ∝ tan θ
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as 1/sin⁴(θ/2).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• N ∝ 1/sin⁴(θ/2) curve shape.

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from ²¹⁴₈₃Bi source directed at thin gold foil (thickness 2.1 × 10^-7 m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles N ∝ 1/sin⁴(θ/2).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ 1/8000 of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles N varies with scattering angle θ as:
  A. N ∝ sin² θ
  B. N ∝ cos² θ
  C. N ∝ 1/sin⁴(θ/2)
  D. N ∝ tan θ
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as 1/sin⁴(θ/2).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• N ∝ 1/sin⁴(θ/2) curve shape.

Experiment Setup

• Performed by Geiger and Marsden under Rutherford's suggestion in 1911.
• 5.5 MeV alpha particles from \(^{214}_{83}Bi\) source directed at thin gold foil (thickness \(2.1 \times 10^{-7}\) m).
• Alpha particles collimated into narrow beam using lead bricks.
• Scattered particles detected by rotatable zinc sulphide screen and microscope.
• Scintillations on ZnS screen indicate alpha particle impacts.

Observations

• Most alpha particles pass through gold foil undeflected → atom mostly empty space.
• Some alpha particles deflect at various angles → presence of small, dense, positively charged nucleus.
• Very few (1 in 8000) deflect by 180° → nucleus size is very small (~1/8000 of atom size).
• Number of scattered particles \(N \propto \frac{1}{\sin^4(\theta/2)}\).

Solved Examples

• Ex 1: What does the deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Calculate the ratio of nucleus size to atom size if 1 in 8000 alpha particles deflect by 180°.
  Answer: Nucleus size ≈ \(\frac{1}{8000}\) of atom size.

Practice Set

• Q1: What does most alpha particles passing through gold foil indicate?
  A. Atom is solid
  B. Atom is mostly empty
  C. Atom has no nucleus
  D. Alpha particles are absorbed
  
• Q2: The scattering angle of alpha particles depends on:
  A. Electron charge
  B. Impact parameter
  C. Alpha particle mass
  D. Foil thickness
  
• Q3: The number of scattered alpha particles \(N\) varies with scattering angle \(\theta\) as:
  A. \(N \propto \sin^2 \theta\)
  B. \(N \propto \cos^2 \theta\)
  C. \(N \propto \frac{1}{\sin^4(\theta/2)}\)
  D. \(N \propto \tan \theta\)
  

Answer Key

• Q1: B - Most alpha particles pass through, showing atom is mostly empty.
• Q2: B - Scattering angle depends on impact parameter (distance from nucleus).
• Q3: C - Number of scattered particles varies as \(1/\sin^4(\theta/2)\).

Memory Aids

• Alpha scattering → nucleus discovery.
• Deflection angle ↔ impact parameter.
• \(N \propto 1/\sin^4(\theta/2)\) curve shape.

Rutherford's Nuclear Model

Postulates

• Positive charge and most mass concentrated in tiny nucleus (~10^-14 m diameter).
• Electrons revolve around nucleus in empty space, held by electrostatic attraction.
• Atom is electrically neutral: total positive charge = total negative charge.

Drawbacks

• Accelerated electrons should radiate energy and spiral into nucleus, collapsing atom (not observed).
• Predicted continuous energy spectrum, but actual atomic spectra are discrete.

Solved Examples

• Ex 1: Why does Rutherford model predict atom collapse?
  Answer: Accelerated electrons radiate energy, lose it, and spiral into nucleus.
• Ex 2: What is the size difference between atom and nucleus?
  Answer: Atom ~10^-10 m, nucleus ~10^-14 m.

Practice Set

• Q1: What is concentrated in the nucleus according to Rutherford?
  A. Electrons
  B. Positive charge and mass
  C. Neutrons only
  D. Empty space
  
• Q2: Why does Rutherford model fail to explain atomic stability?
  A. Electrons do not revolve
  B. Electrons radiate energy and collapse
  C. Nucleus is too big
  D. Atoms are not neutral
  
• Q3: What type of spectrum does Rutherford model fail to explain?
  A. Continuous
  B. Discrete
  C. Emission
  D. Absorption
  

Answer Key

• Q1: B - Positive charge and mass are concentrated in nucleus.
• Q2: B - Accelerated electrons radiate energy and collapse atom.
• Q3: A - Model predicts continuous spectrum, but actual is discrete.

Memory Aids

• Nucleus = dense positive core.
• Electron spiral collapse = drawback.
• Atom mostly empty space.

Postulates

• Positive charge and most mass concentrated in tiny nucleus (~10^-14 m diameter).
• Electrons revolve around nucleus in empty space, held by electrostatic attraction.
• Atom is electrically neutral: total positive charge = total negative charge.

Drawbacks

• Accelerated electrons should radiate energy and spiral into nucleus, collapsing atom (not observed).
• Predicted continuous energy spectrum, but actual atomic spectra are discrete.

Solved Examples

• Ex 1: Why does Rutherford model predict atom collapse?
  Answer: Accelerated electrons radiate energy, lose it, and spiral into nucleus.
• Ex 2: What is the size difference between atom and nucleus?
  Answer: Atom ~10^-10 m, nucleus ~10^-14 m.

Practice Set

• Q1: What is concentrated in the nucleus according to Rutherford?
  A. Electrons
  B. Positive charge and mass
  C. Neutrons only
  D. Empty space
  
• Q2: Why does Rutherford model fail to explain atomic stability?
  A. Electrons do not revolve
  B. Electrons radiate energy and collapse
  C. Nucleus is too big
  D. Atoms are not neutral
  
• Q3: What type of spectrum does Rutherford model fail to explain?
  A. Continuous
  B. Discrete
  C. Emission
  D. Absorption
  

Answer Key

• Q1: B - Positive charge and mass are concentrated in nucleus.
• Q2: B - Accelerated electrons radiate energy and collapse atom.
• Q3: A - Model predicts continuous spectrum, but actual is discrete.

Memory Aids

• Nucleus = dense positive core.
• Electron spiral collapse = drawback.
• Atom mostly empty space.

Postulates

• Positive charge and most mass concentrated in tiny nucleus (~\(10^{-14}\) m diameter).
• Electrons revolve around nucleus in empty space, held by electrostatic attraction.
• Atom is electrically neutral: total positive charge = total negative charge.

Drawbacks

• Accelerated electrons should radiate energy and spiral into nucleus, collapsing atom (not observed).
• Predicted continuous energy spectrum, but actual atomic spectra are discrete.

Solved Examples

• Ex 1: Why does Rutherford model predict atom collapse?
  Answer: Accelerated electrons radiate energy, lose it, and spiral into nucleus.
• Ex 2: What is the size difference between atom and nucleus?
  Answer: Atom ~\(10^{-10}\) m, nucleus ~\(10^{-14}\) m.

Practice Set

• Q1: What is concentrated in the nucleus according to Rutherford?
  A. Electrons
  B. Positive charge and mass
  C. Neutrons only
  D. Empty space
  
• Q2: Why does Rutherford model fail to explain atomic stability?
  A. Electrons do not revolve
  B. Electrons radiate energy and collapse
  C. Nucleus is too big
  D. Atoms are not neutral
  
• Q3: What type of spectrum does Rutherford model fail to explain?
  A. Continuous
  B. Discrete
  C. Emission
  D. Absorption
  

Answer Key

• Q1: B - Positive charge and mass are concentrated in nucleus.
• Q2: B - Accelerated electrons radiate energy and collapse atom.
• Q3: A - Model predicts continuous spectrum, but actual is discrete.

Memory Aids

• Nucleus = dense positive core.
• Electron spiral collapse = drawback.
• Atom mostly empty space.

Electron Orbits

Forces and Energy

• Electrostatic force F_e provides centripetal force F_c for electron orbit.
• F_e = F_c ⇒ 1/(4π ε₀) e²/r² = mv²/r
• Orbit radius: r = e²/(4π ε₀mv²)
• Kinetic energy: K = 1/2 mv² = e²/(8π ε₀r)
• Potential energy: U = - e²/(4π ε₀r)
• Total energy: E = K + U = - e²/(8π ε₀r) (negative, electron bound to nucleus)

Solved Examples

• Ex 1: Calculate orbit radius if electron velocity v is known.
  Use r = e²/(4π ε₀mv²).
• Ex 2: Find total energy of electron in orbit radius r.
  Use E = - e²/(8π ε₀r).

Practice Set

• Q1: What force keeps electron in orbit?
  A. Gravitational
  B. Electrostatic attraction
  C. Magnetic
  D. Nuclear
  
• Q2: Total energy of electron in orbit is:
  A. Positive
  B. Zero
  C. Negative
  D. Infinite
  
• Q3: If electron velocity increases, orbit radius:
  A. Increases
  B. Decreases
  C. Remains same
  D. Becomes zero
  

Answer Key

• Q1: B - Electrostatic force provides centripetal force.
• Q2: C - Total energy is negative, electron is bound.
• Q3: B - Radius inversely proportional to square of velocity.

Memory Aids

• Orbit radius r ∝ 1/v².
• Total energy negative → bound electron.
• Electrostatic force = centripetal force.

Forces and Energy

• Electrostatic force F_e provides centripetal force F_c for electron orbit.
• F_e = F_c ⇒ 1/(4π ε₀) e²/r² = mv²/r
• Orbit radius: r = e²/(4π ε₀mv²)
• Kinetic energy: K = 1/2 mv² = e²/(8π ε₀r)
• Potential energy: U = - e²/(4π ε₀r)
• Total energy: E = K + U = - e²/(8π ε₀r) (negative, electron bound to nucleus)

Solved Examples

• Ex 1: Calculate orbit radius if electron velocity v is known.
  Use r = e²/(4π ε₀mv²).
• Ex 2: Find total energy of electron in orbit radius r.
  Use E = - e²/(8π ε₀r).

Practice Set

• Q1: What force keeps electron in orbit?
  A. Gravitational
  B. Electrostatic attraction
  C. Magnetic
  D. Nuclear
  
• Q2: Total energy of electron in orbit is:
  A. Positive
  B. Zero
  C. Negative
  D. Infinite
  
• Q3: If electron velocity increases, orbit radius:
  A. Increases
  B. Decreases
  C. Remains same
  D. Becomes zero
  

Answer Key

• Q1: B - Electrostatic force provides centripetal force.
• Q2: C - Total energy is negative, electron is bound.
• Q3: B - Radius inversely proportional to square of velocity.

Memory Aids

• Orbit radius r ∝ 1/v².
• Total energy negative → bound electron.
• Electrostatic force = centripetal force.

Forces and Energy

• Electrostatic force \(F_e\) provides centripetal force \(F_c\) for electron orbit.
• \(F_e = F_c \Rightarrow \frac{1}{4\pi \epsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}\)
• Orbit radius: \(r = \frac{e^2}{4\pi \epsilon_0 mv^2}\)
• Kinetic energy: \(K = \frac{1}{2} mv^2 = \frac{e^2}{8\pi \epsilon_0 r}\)
• Potential energy: \(U = - \frac{e^2}{4\pi \epsilon_0 r}\)
• Total energy: \(E = K + U = - \frac{e^2}{8\pi \epsilon_0 r}\) (negative, electron bound to nucleus)

Solved Examples

• Ex 1: Calculate orbit radius if electron velocity \(v\) is known.
  Use \(r = \frac{e^2}{4\pi \epsilon_0 mv^2}\).
• Ex 2: Find total energy of electron in orbit radius \(r\).
  Use \(E = - \frac{e^2}{8\pi \epsilon_0 r}\).

Practice Set

• Q1: What force keeps electron in orbit?
  A. Gravitational
  B. Electrostatic attraction
  C. Magnetic
  D. Nuclear
  
• Q2: Total energy of electron in orbit is:
  A. Positive
  B. Zero
  C. Negative
  D. Infinite
  
• Q3: If electron velocity increases, orbit radius:
  A. Increases
  B. Decreases
  C. Remains same
  D. Becomes zero
  

Answer Key

• Q1: B - Electrostatic force provides centripetal force.
• Q2: C - Total energy is negative, electron is bound.
• Q3: B - Radius inversely proportional to square of velocity.

Memory Aids

• Orbit radius \(r \propto \frac{1}{v^2}\).
• Total energy negative → bound electron.
• Electrostatic force = centripetal force.

Atomic Spectra

Emission Line Spectrum

• Excited atomic gas emits radiation at specific wavelengths only.
• Emission line spectrum: bright lines on dark background.
• Example: Hydrogen emission spectrum shows distinct lines.

Explanation of Hydrogen Spectrum

• Electrons change energy levels, emitting light at specific wavelengths.
• Series: Lyman (UV), Balmer (visible), Paschen (IR).
• Lyman: 91-122 nm, Balmer: 365-656 nm, Paschen: 820-1875 nm.
• Each line = electron dropping to lower energy level, releasing photon.

Solved Examples

• Ex 1: Identify the series for a line at 486.1 nm.
  Solution: 486.1 nm lies in visible range → Balmer series.

Practice Set

• Q1: Which series of hydrogen spectrum lies in the ultraviolet region?
  A. Balmer
  B. Paschen
  C. Lyman
  D. Brackett
• Q2: The emission line at 656.3 nm belongs to which series?
  A. Lyman
  B. Balmer
  C. Paschen
  D. Pfund
• Q3: What happens when an electron falls from n=4 to n=2 in hydrogen?
  A. Emits photon in Balmer series
  B. Emits photon in Lyman series
  C. Absorbs photon
  D. Emits photon in Paschen series

Answer Key

• Q1: C - Lyman series is ultraviolet.
• Q2: B - 656.3 nm is Hα line in Balmer series.
• Q3: A - Transition from n=4 to n=2 emits Balmer series photon.

Memory Aids

• Lyman: UV, n=1 final level.
• Balmer: Visible, n=2 final level.
• Paschen: IR, n=3 final level.

Emission Line Spectrum

• Excited atomic gas emits radiation at specific wavelengths only.
• Emission line spectrum: bright lines on dark background.
• Example: Hydrogen emission spectrum shows distinct lines.

Explanation of Hydrogen Spectrum

• Electrons change energy levels, emitting light at specific wavelengths.
• Series: Lyman (UV), Balmer (visible), Paschen (IR).
• Lyman: 91-122 nm, Balmer: 365-656 nm, Paschen: 820-1875 nm.
• Each line = electron dropping to lower energy level, releasing photon.

Solved Examples

• Ex 1: Identify the series for a line at 486.1 nm.
  Solution: 486.1 nm lies in visible range → Balmer series.

Practice Set

• Q1: Which series of hydrogen spectrum lies in the ultraviolet region?
  A. Balmer
  B. Paschen
  C. Lyman
  D. Brackett
• Q2: The emission line at 656.3 nm belongs to which series?
  A. Lyman
  B. Balmer
  C. Paschen
  D. Pfund
• Q3: What happens when an electron falls from n=4 to n=2 in hydrogen?
  A. Emits photon in Balmer series
  B. Emits photon in Lyman series
  C. Absorbs photon
  D. Emits photon in Paschen series

Answer Key

• Q1: C - Lyman series is ultraviolet.
• Q2: B - 656.3 nm is Hα line in Balmer series.
• Q3: A - Transition from n=4 to n=2 emits Balmer series photon.

Memory Aids

• Lyman: UV, n=1 final level.
• Balmer: Visible, n=2 final level.
• Paschen: IR, n=3 final level.

Emission Line Spectrum

• Excited atomic gas emits radiation at specific wavelengths only.
• Emission line spectrum: bright lines on dark background.
• Example: Hydrogen emission spectrum shows distinct lines.

Explanation of Hydrogen Spectrum

• Electrons change energy levels, emitting light at specific wavelengths.
• Series: Lyman (UV), Balmer (visible), Paschen (IR).
• Lyman: 91-122 nm, Balmer: 365-656 nm, Paschen: 820-1875 nm.
• Each line = electron dropping to lower energy level, releasing photon.

Solved Examples

• Ex 1: Identify the series for a line at 486.1 nm.
  Solution: 486.1 nm lies in visible range → Balmer series.

Practice Set

• Q1: Which series of hydrogen spectrum lies in the ultraviolet region?
  A. Balmer
  B. Paschen
  C. Lyman
  D. Brackett
• Q2: The emission line at 656.3 nm belongs to which series?
  A. Lyman
  B. Balmer
  C. Paschen
  D. Pfund
• Q3: What happens when an electron falls from n=4 to n=2 in hydrogen?
  A. Emits photon in Balmer series
  B. Emits photon in Lyman series
  C. Absorbs photon
  D. Emits photon in Paschen series

Answer Key

• Q1: C - Lyman series is ultraviolet.
• Q2: B - 656.3 nm is Hα line in Balmer series.
• Q3: A - Transition from n=4 to n=2 emits Balmer series photon.

Memory Aids

• Lyman: UV, n=1 final level.
• Balmer: Visible, n=2 final level.
• Paschen: IR, n=3 final level.

Absorption Spectrum

Key Points

• White light passing through gas shows dark lines at specific wavelengths.
• Dark lines match emission lines of the gas.
• Called absorption spectrum of the gas.

Solved Examples

• Ex 1: Why do dark lines appear in absorption spectrum?
  Solution: Gas atoms absorb light at specific wavelengths, causing dark lines.

Practice Set

• Q1: What causes dark lines in absorption spectrum?
  A. Emission of light
  B. Absorption of specific wavelengths
  C. Scattering of light
  D. Reflection of light
• Q2: Absorption lines correspond to which of the following?
  A. Random wavelengths
  B. Emission lines of the gas
  C. Continuous spectrum
  D. Infrared radiation only
• Q3: Absorption spectrum is observed when white light passes through?
  A. Vacuum
  B. Gas
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Gas absorbs specific wavelengths causing dark lines.
• Q2: B - Absorption lines match emission lines of the gas.
• Q3: B - Absorption spectrum occurs when light passes through gas.

Memory Aids

• Absorption = dark lines, emission = bright lines.
• Absorption lines match emission lines of same gas.

Key Points

• White light passing through gas shows dark lines at specific wavelengths.
• Dark lines match emission lines of the gas.
• Called absorption spectrum of the gas.

Solved Examples

• Ex 1: Why do dark lines appear in absorption spectrum?
  Solution: Gas atoms absorb light at specific wavelengths, causing dark lines.

Practice Set

• Q1: What causes dark lines in absorption spectrum?
  A. Emission of light
  B. Absorption of specific wavelengths
  C. Scattering of light
  D. Reflection of light
• Q2: Absorption lines correspond to which of the following?
  A. Random wavelengths
  B. Emission lines of the gas
  C. Continuous spectrum
  D. Infrared radiation only
• Q3: Absorption spectrum is observed when white light passes through?
  A. Vacuum
  B. Gas
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Gas absorbs specific wavelengths causing dark lines.
• Q2: B - Absorption lines match emission lines of the gas.
• Q3: B - Absorption spectrum occurs when light passes through gas.

Memory Aids

• Absorption = dark lines, emission = bright lines.
• Absorption lines match emission lines of same gas.

Key Points

• White light passing through gas shows dark lines at specific wavelengths.
• Dark lines match emission lines of the gas.
• Called absorption spectrum of the gas.

Solved Examples

• Ex 1: Why do dark lines appear in absorption spectrum?
  Solution: Gas atoms absorb light at specific wavelengths, causing dark lines.

Practice Set

• Q1: What causes dark lines in absorption spectrum?
  A. Emission of light
  B. Absorption of specific wavelengths
  C. Scattering of light
  D. Reflection of light
• Q2: Absorption lines correspond to which of the following?
  A. Random wavelengths
  B. Emission lines of the gas
  C. Continuous spectrum
  D. Infrared radiation only
• Q3: Absorption spectrum is observed when white light passes through?
  A. Vacuum
  B. Gas
  C. Solid
  D. Liquid

Answer Key

• Q1: B - Gas absorbs specific wavelengths causing dark lines.
• Q2: B - Absorption lines match emission lines of the gas.
• Q3: B - Absorption spectrum occurs when light passes through gas.

Memory Aids

• Absorption = dark lines, emission = bright lines.
• Absorption lines match emission lines of same gas.

Spectral Series

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,...
• R = Rydberg constant = 1.097 × 10⁷ m^-1

Spectral Series of Hydrogen

• Lyman: 1/λ = R ( 1/1² - 1/n² ), n=2,3,4,... (UV)
• Balmer: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,... (Visible)
• Paschen: 1/λ = R ( 1/3² - 1/n² ), n=4,5,6,... (IR)
• Brackett: 1/λ = R ( 1/4² - 1/n² ), n=5,6,7,... (IR)
• Pfund: 1/λ = R ( 1/5² - 1/n² ), n=6,7,8,... (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with n=3.
  Solution: 1/λ = R (1/2² - 1/3²) = 1.097 × 10⁷ (1/4 - 1/9) = 1.52 × 10⁶ m^-1
  λ = 6.58 × 10^-7 m = 658 nm

Practice Set

• Q1: Which series corresponds to electron transition to n=1?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. 1.097 × 10⁷ m^-1
  B. 9.8 m/s²
  C. 3 × 10⁸ m/s
  D. 6.63 × 10^-34 Js
• Q3: Wavelength of Paschen series line for transition n=5 → n=3 is?
  A. Calculate using 1/λ = R (1/3² - 1/5²)
  B. 656.3 nm
  C. 486.1 nm
  D. 91.2 nm

Answer Key

• Q1: B - Lyman series ends at n=1.
• Q2: A - Rydberg constant is 1.097 × 10⁷ m^-1.
• Q3: A - Use formula 1/λ = R (1/3² - 1/5²) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: 1/λ = R (1/n_f² - 1/n_i²).

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,...
• R = Rydberg constant = 1.097 × 10⁷ m^-1

Spectral Series of Hydrogen

• Lyman: 1/λ = R ( 1/1² - 1/n² ), n=2,3,4,... (UV)
• Balmer: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,... (Visible)
• Paschen: 1/λ = R ( 1/3² - 1/n² ), n=4,5,6,... (IR)
• Brackett: 1/λ = R ( 1/4² - 1/n² ), n=5,6,7,... (IR)
• Pfund: 1/λ = R ( 1/5² - 1/n² ), n=6,7,8,... (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with n=3.
  Solution: 1/λ = R (1/2² - 1/3²) = 1.097 × 10⁷ (1/4 - 1/9) = 1.52 × 10⁶ m^-1
  λ = 6.58 × 10^-7 m = 658 nm

Practice Set

• Q1: Which series corresponds to electron transition to n=1?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. 1.097 × 10⁷ m^-1
  B. 9.8 m/s²
  C. 3 × 10⁸ m/s
  D. 6.63 × 10^-34 Js
• Q3: Wavelength of Paschen series line for transition n=5 → n=3 is?
  A. Calculate using 1/λ = R (1/3² - 1/5²)
  B. 656.3 nm
  C. 486.1 nm
  D. 91.2 nm

Answer Key

• Q1: B - Lyman series ends at n=1.
• Q2: A - Rydberg constant is 1.097 × 10⁷ m^-1.
• Q3: A - Use formula 1/λ = R (1/3² - 1/5²) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: 1/λ = R (1/n_f² - 1/n_i²).

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,...
• R = Rydberg constant = 1.097 × 10⁷ m^-1

Spectral Series of Hydrogen

• Lyman: 1/λ = R ( 1/1² - 1/n² ), n=2,3,4,... (UV)
• Balmer: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,... (Visible)
• Paschen: 1/λ = R ( 1/3² - 1/n² ), n=4,5,6,... (IR)
• Brackett: 1/λ = R ( 1/4² - 1/n² ), n=5,6,7,... (IR)
• Pfund: 1/λ = R ( 1/5² - 1/n² ), n=6,7,8,... (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with n=3.
  Solution: 1/λ = R (1/2² - 1/3²) = 1.097 × 10⁷ (1/4 - 1/9) = 1.52 × 10⁶ m^-1
  λ = 6.58 × 10^-7 m = 658 nm

Practice Set

• Q1: Which series corresponds to electron transition to n=1?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. 1.097 × 10⁷ m^-1
  B. 9.8 m/s²
  C. 3 × 10⁸ m/s
  D. 6.63 × 10^-34 Js
• Q3: Wavelength of Paschen series line for transition n=5 → n=3 is?
  A. Calculate using 1/λ = R (1/3² - 1/5²)
  B. 656.3 nm
  C. 486.1 nm
  D. 91.2 nm

Answer Key

• Q1: B - Lyman series ends at n=1.
• Q2: A - Rydberg constant is 1.097 × 10⁷ m^-1.
• Q3: A - Use formula 1/λ = R (1/3² - 1/5²) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: 1/λ = R (1/n_f² - 1/n_i²).

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,...
• R = Rydberg constant = 1.097 × 10⁷ m^-1

Spectral Series of Hydrogen

• Lyman: 1/λ = R ( 1/1² - 1/n² ), n=2,3,4,... (UV)
• Balmer: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,... (Visible)
• Paschen: 1/λ = R ( 1/3² - 1/n² ), n=4,5,6,... (IR)
• Brackett: 1/λ = R ( 1/4² - 1/n² ), n=5,6,7,... (IR)
• Pfund: 1/λ = R ( 1/5² - 1/n² ), n=6,7,8,... (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with n=3.
  Solution: 1/λ = R (1/2² - 1/3²) = 1.097 × 10⁷ (1/4 - 1/9) = 1.52 × 10⁶ m^-1
  λ = 6.58 × 10^-7 m = 658 nm

Practice Set

• Q1: Which series corresponds to electron transition to n=1?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. 1.097 × 10⁷ m^-1
  B. 9.8 m/s²
  C. 3 × 10⁸ m/s
  D. 6.63 × 10^-34 Js
• Q3: Wavelength of Paschen series line for transition n=5 → n=3 is?
  A. Calculate using 1/λ = R (1/3² - 1/5²)
  B. 656.3 nm
  C. 486.1 nm
  D. 91.2 nm

Answer Key

• Q1: B - Lyman series ends at n=1.
• Q2: A - Rydberg constant is 1.097 × 10⁷ m^-1.
• Q3: A - Use formula 1/λ = R (1/3² - 1/5²) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: 1/λ = R (1/n_f² - 1/n_i²).

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,...
• R = Rydberg constant = 1.097 × 10⁷ m^-1

Spectral Series of Hydrogen

• Lyman: 1/λ = R ( 1/1² - 1/n² ), n=2,3,4,... (UV)
• Balmer: 1/λ = R ( 1/2² - 1/n² ), n=3,4,5,... (Visible)
• Paschen: 1/λ = R ( 1/3² - 1/n² ), n=4,5,6,... (IR)
• Brackett: 1/λ = R ( 1/4² - 1/n² ), n=5,6,7,... (IR)
• Pfund: 1/λ = R ( 1/5² - 1/n² ), n=6,7,8,... (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with n=3.
  Solution: 1/λ = R (1/2² - 1/3²) = 1.097 × 10⁷ (1/4 - 1/9) = 1.52 × 10⁶ m^-1
  λ = 6.58 × 10^-7 m = 658 nm

Practice Set

• Q1: Which series corresponds to electron transition to n=1?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. 1.097 × 10⁷ m^-1
  B. 9.8 m/s²
  C. 3 × 10⁸ m/s
  D. 6.63 × 10^-34 Js
• Q3: Wavelength of Paschen series line for transition n=5 → n=3 is?
  A. Calculate using 1/λ = R (1/3² - 1/5²)
  B. 656.3 nm
  C. 486.1 nm
  D. 91.2 nm

Answer Key

• Q1: B - Lyman series ends at n=1.
• Q2: A - Rydberg constant is 1.097 × 10⁷ m^-1.
• Q3: A - Use formula 1/λ = R (1/3² - 1/5²) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: 1/λ = R (1/n_f² - 1/n_i²).

Definition and Pattern

• Light frequencies emitted by elements show regular patterns called spectral series.
• Hydrogen has simplest spectrum with sets of lines called series.
• Balmer series observed in visible region (1885).
• Lines get closer and weaker as wavelength decreases.
• Balmer formula: \( \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \), \( n=3,4,5,... \)
• R = Rydberg constant = \(1.097 \times 10^7 \text{ m}^{-1}\)

Spectral Series of Hydrogen

• Lyman: \( \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right), n=2,3,4,... \) (UV)
• Balmer: \( \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right), n=3,4,5,... \) (Visible)
• Paschen: \( \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n^2} \right), n=4,5,6,... \) (IR)
• Brackett: \( \frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} \right), n=5,6,7,... \) (IR)
• Pfund: \( \frac{1}{\lambda} = R \left( \frac{1}{5^2} - \frac{1}{n^2} \right), n=6,7,8,... \) (IR)

Solved Examples

• Ex 1: Calculate wavelength for Balmer line with \( n=3 \).
  Solution: \( \frac{1}{\lambda} = R (\frac{1}{2^2} - \frac{1}{3^2}) = 1.097 \times 10^7 (\frac{1}{4} - \frac{1}{9}) = 1.52 \times 10^6 \text{ m}^{-1} \)
  \( \lambda = 6.58 \times 10^{-7} \text{ m} = 658 \text{ nm} \)

Practice Set

• Q1: Which series corresponds to electron transition to \( n=1 \)?
  A. Balmer
  B. Lyman
  C. Paschen
  D. Brackett
• Q2: The Rydberg constant value is?
  A. \(1.097 \times 10^7 \text{ m}^{-1}\)
  B. \(9.8 \text{ m/s}^2\)
  C. \(3 \times 10^8 \text{ m/s}\)
  D. \(6.63 \times 10^{-34} \text{ Js}\)
• Q3: Wavelength of Paschen series line for transition \( n=5 \to n=3 \) is?
  A. Calculate using \( \frac{1}{\lambda} = R (\frac{1}{3^2} - \frac{1}{5^2}) \)
  B. \( 656.3 \text{ nm} \)
  C. \( 486.1 \text{ nm} \)
  D. \( 91.2 \text{ nm} \)

Answer Key

• Q1: B - Lyman series ends at \( n=1 \).
• Q2: A - Rydberg constant is \(1.097 \times 10^7 \text{ m}^{-1}\).
• Q3: A - Use formula \( \frac{1}{\lambda} = R (\frac{1}{3^2} - \frac{1}{5^2}) \) to calculate.

Memory Aids

• Series final levels: Lyman=1, Balmer=2, Paschen=3.
• Balmer visible, Lyman UV, others IR.
• Formula: \( \frac{1}{\lambda} = R (\frac{1}{n_f^2} - \frac{1}{n_i^2}) \).

Bohr Model

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: mvr = n h/2π, where n = 1, 2, 3, ... .
• Electron transitions between orbits emit or absorb energy as photons: hν = E_i - E_f.
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for n=2: L = 2 × h/2π = 2h/2π.
• Ex 2: Energy of photon emitted when electron falls from n=3 to n=2: hν = E₃ - E₂.

Practice Set

• Q1: What is the angular momentum of electron in n=1 orbit?
  • A) h/2π
  • B) h
  • C) 2h
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in n=3 orbit has angular momentum:
  • A) 3 h/2π
  • B) h/2π
  • C) 2 h/2π
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is n h/2π, for n=1 it is h/2π.
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of h/2π, here n=3.

Memory Aids

• Angular momentum = n h/2π
• Energy emitted/absorbed = hν = E_i - E_f
• Electron orbits are stable and quantized

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: mvr = n h/2π, where n = 1, 2, 3, ... .
• Electron transitions between orbits emit or absorb energy as photons: hν = E_i - E_f.
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for n=2: L = 2 × h/2π = 2h/2π.
• Ex 2: Energy of photon emitted when electron falls from n=3 to n=2: hν = E₃ - E₂.

Practice Set

• Q1: What is the angular momentum of electron in n=1 orbit?
  • A) h/2π
  • B) h
  • C) 2h
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in n=3 orbit has angular momentum:
  • A) 3 h/2π
  • B) h/2π
  • C) 2 h/2π
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is n h/2π, for n=1 it is h/2π.
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of h/2π, here n=3.

Memory Aids

• Angular momentum = n h/2π
• Energy emitted/absorbed = hν = E_i - E_f
• Electron orbits are stable and quantized

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: mvr = n h/2π, where n = 1, 2, 3, ... .
• Electron transitions between orbits emit or absorb energy as photons: hν = E_i - E_f.
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for n=2: L = 2 × h/2π = 2h/2π.
• Ex 2: Energy of photon emitted when electron falls from n=3 to n=2: hν = E₃ - E₂.

Practice Set

• Q1: What is the angular momentum of electron in n=1 orbit?
  • A) h/2π
  • B) h
  • C) 2h
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in n=3 orbit has angular momentum:
  • A) 3 h/2π
  • B) h/2π
  • C) 2 h/2π
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is n h/2π, for n=1 it is h/2π.
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of h/2π, here n=3.

Memory Aids

• Angular momentum = n h/2π
• Energy emitted/absorbed = hν = E_i - E_f
• Electron orbits are stable and quantized

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: mvr = n h/2π, where n = 1, 2, 3, ... .
• Electron transitions between orbits emit or absorb energy as photons: hν = E_i - E_f.
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for n=2: L = 2 × h/2π = 2h/2π.
• Ex 2: Energy of photon emitted when electron falls from n=3 to n=2: hν = E₃ - E₂.

Practice Set

• Q1: What is the angular momentum of electron in n=1 orbit?
  • A) h/2π
  • B) h
  • C) 2h
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in n=3 orbit has angular momentum:
  • A) 3 h/2π
  • B) h/2π
  • C) 2 h/2π
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is n h/2π, for n=1 it is h/2π.
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of h/2π, here n=3.

Memory Aids

• Angular momentum = n h/2π
• Energy emitted/absorbed = hν = E_i - E_f
• Electron orbits are stable and quantized

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: mvr = n h/2π, where n = 1, 2, 3, ... .
• Electron transitions between orbits emit or absorb energy as photons: hν = E_i - E_f.
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for n=2: L = 2 × h/2π = 2h/2π.
• Ex 2: Energy of photon emitted when electron falls from n=3 to n=2: hν = E₃ - E₂.

Practice Set

• Q1: What is the angular momentum of electron in n=1 orbit?
  • A) h/2π
  • B) h
  • C) 2h
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in n=3 orbit has angular momentum:
  • A) 3 h/2π
  • B) h/2π
  • C) 2 h/2π
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is n h/2π, for n=1 it is h/2π.
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of h/2π, here n=3.

Memory Aids

• Angular momentum = n h/2π
• Energy emitted/absorbed = hν = E_i - E_f
• Electron orbits are stable and quantized

Bohr's Postulates

• Electrons revolve in fixed stable orbits without emitting energy.
• Angular momentum of electron in orbit: \( mvr = n \frac{h}{2\pi} \), where \( n = 1, 2, 3, \ldots \).
• Electron transitions between orbits emit or absorb energy as photons: \( h\nu = E_i - E_f \).
• Energy quantization: electrons have discrete energy levels.

Solved Examples

• Ex 1: Calculate angular momentum for \( n=2 \): \( L = 2 \times \frac{h}{2\pi} = \frac{2h}{2\pi} \).
• Ex 2: Energy of photon emitted when electron falls from \( n=3 \) to \( n=2 \): \( h\nu = E_3 - E_2 \).

Practice Set

• Q1: What is the angular momentum of electron in \( n=1 \) orbit?
  • A) \( \frac{h}{2\pi} \)
  • B) \( h \)
  • C) \( 2h \)
  • D) Zero
• Q2: When electron jumps from higher to lower orbit, it:
  • A) Absorbs energy
  • B) Emits energy as photon
  • C) Remains same energy
  • D) Emits sound
• Q3: Electron in \( n=3 \) orbit has angular momentum:
  • A) \( 3 \frac{h}{2\pi} \)
  • B) \( \frac{h}{2\pi} \)
  • C) \( 2 \frac{h}{2\pi} \)
  • D) Zero

Answer Key

• Q1: A - By Bohr's postulate, angular momentum is \( n \frac{h}{2\pi} \), for \( n=1 \) it is \( \frac{h}{2\pi} \).
• Q2: B - Electron emits photon energy when falling to lower orbit.
• Q3: A - Angular momentum is integral multiple of \( \frac{h}{2\pi} \), here \( n=3 \).

Memory Aids

• Angular momentum \( = n \frac{h}{2\pi} \)
• Energy emitted/absorbed \( = h\nu = E_i - E_f \)
• Electron orbits are stable and quantized

Energy Transitions

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy E_i to lower E_f orbit emitting photon energy hν = E_i - E_f.
• Photon frequency ν relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from E₃ = -1.51 eV to E₂ = -3.4 eV: hν = |-1.51 - (-3.4)| = 1.89 eV.

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) E_f - E_i
  • B) E_i - E_f
  • C) E_i + E_f
  • D) Zero
• Q3: Electron falls from n=4 to n=2, energy emitted is:
  • A) 0.85 - 3.4 eV
  • B) 3.4 - 0.85 eV
  • C) |-0.85 - (-3.4)| eV
  • D) 0 eV

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference E_i - E_f.
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = hν = E_i - E_f
• Photon frequency ν proportional to energy difference
• Electron jumps cause emission or absorption

Electron Orbit Transitions

• Electron moves from higher energy \( E_i \) to lower \( E_f \) orbit emitting photon energy \( h\nu = E_i - E_f \).
• Photon frequency \( \nu \) relates to energy difference.
• Energy levels are discrete and quantized.

Solved Examples

• Ex 1: Calculate photon energy when electron falls from \( E_3 = -1.51 \text{ eV} \) to \( E_2 = -3.4 \text{ eV} \): \( h\nu = |-1.51 - (-3.4)| = 1.89 \text{ eV} \).

Practice Set

• Q1: What happens when electron jumps from higher to lower orbit?
  • A) Absorbs photon
  • B) Emits photon
  • C) No energy change
  • D) Emits electron
• Q2: Energy of photon emitted equals:
  • A) \( E_f - E_i \)
  • B) \( E_i - E_f \)
  • C) \( E_i + E_f \)
  • D) Zero
• Q3: Electron falls from \( n=4 \) to \( n=2 \), energy emitted is:
  • A) \( 0.85 - 3.4 \text{ eV} \)
  • B) \( 3.4 - 0.85 \text{ eV} \)
  • C) \( |-0.85 - (-3.4)| \text{ eV} \)
  • D) \( 0 \text{ eV} \)

Answer Key

• Q1: B - Electron emits photon when moving to lower orbit.
• Q2: B - Photon energy equals difference \( E_i - E_f \).
• Q3: C - Energy emitted is difference of energy levels.

Memory Aids

• Energy emitted = \( h\nu = E_i - E_f \)
• Photon frequency \( \nu \) proportional to energy difference
• Electron jumps cause emission or absorption

Orbit Radius & Energy Levels

Formulas and Energy Levels

• Centripetal force: mv_n²/r_n = 1/(4πϵ₀) · eZe/r_n²
• Angular momentum: L_n = m v_n r_n = n h/2π
• Radius of nth orbit: r_n = n² h² ϵ₀/(π m e² Z) = 0.53 n²/Z Å
• Energy levels: E_n = -13.6/n² eV for hydrogen atom (Z=1)
• Ground state (n=1) energy: -13.6 eV, higher states have less negative energy.

Solved Examples

• Ex 1: Calculate radius for n=2 in hydrogen: r₂ = 0.53 × 2²/1 = 2.12 Å.
• Ex 2: Energy at n=3: E₃ = -13.6/3² = -1.51 eV.

Practice Set

• Q1: Radius of n=1 orbit in hydrogen is:
  • A) 0.53 Å
  • B) 1.06 Å
  • C) 2.12 Å
  • D) 0.26 Å
• Q2: Energy of electron in n=4 orbit is:
  • A) -0.85 eV
  • B) -3.4 eV
  • C) -13.6 eV
  • D) -1.51 eV
• Q3: Radius r_n is proportional to:
  • A) n
  • B) n²/Z
  • C) Z/n²
  • D) 1/n

Answer Key

• Q1: A - Bohr radius for hydrogen ground state is 0.53 Å.
• Q2: A - Energy at n=4 is -13.6/4² = -0.85 eV.
• Q3: B - Radius varies as n²/Z.

Memory Aids

• Radius r_n = 0.53 n²/Z Å
• Energy E_n = -13.6/n² eV
• Ground state energy = -13.6 eV

Formulas and Energy Levels

• Centripetal force: mv_n²/r_n = 1/(4πϵ₀) · eZe/r_n²
• Angular momentum: L_n = m v_n r_n = n h/2π
• Radius of nth orbit: r_n = n² h² ϵ₀/(π m e² Z) = 0.53 n²/Z Å
• Energy levels: E_n = -13.6/n² eV for hydrogen atom (Z=1)
• Ground state (n=1) energy: -13.6 eV, higher states have less negative energy.

Solved Examples

• Ex 1: Calculate radius for n=2 in hydrogen: r₂ = 0.53 × 2²/1 = 2.12 Å.
• Ex 2: Energy at n=3: E₃ = -13.6/3² = -1.51 eV.

Practice Set

• Q1: Radius of n=1 orbit in hydrogen is:
  • A) 0.53 Å
  • B) 1.06 Å
  • C) 2.12 Å
  • D) 0.26 Å
• Q2: Energy of electron in n=4 orbit is:
  • A) -0.85 eV
  • B) -3.4 eV
  • C) -13.6 eV
  • D) -1.51 eV
• Q3: Radius r_n is proportional to:
  • A) n
  • B) n²/Z
  • C) Z/n²
  • D) 1/n

Answer Key

• Q1: A - Bohr radius for hydrogen ground state is 0.53 Å.
• Q2: A - Energy at n=4 is -13.6/4² = -0.85 eV.
• Q3: B - Radius varies as n²/Z.

Memory Aids

• Radius r_n = 0.53 n²/Z Å
• Energy E_n = -13.6/n² eV
• Ground state energy = -13.6 eV

Formulas and Energy Levels

• Centripetal force: mv_n²/r_n = 1/(4πϵ₀) · eZe/r_n²
• Angular momentum: L_n = m v_n r_n = n h/2π
• Radius of nth orbit: r_n = n² h² ϵ₀/(π m e² Z) = 0.53 n²/Z Å
• Energy levels: E_n = -13.6/n² eV for hydrogen atom (Z=1)
• Ground state (n=1) energy: -13.6 eV, higher states have less negative energy.

Solved Examples

• Ex 1: Calculate radius for n=2 in hydrogen: r₂ = 0.53 × 2²/1 = 2.12 Å.
• Ex 2: Energy at n=3: E₃ = -13.6/3² = -1.51 eV.

Practice Set

• Q1: Radius of n=1 orbit in hydrogen is:
  • A) 0.53 Å
  • B) 1.06 Å
  • C) 2.12 Å
  • D) 0.26 Å
• Q2: Energy of electron in n=4 orbit is:
  • A) -0.85 eV
  • B) -3.4 eV
  • C) -13.6 eV
  • D) -1.51 eV
• Q3: Radius r_n is proportional to:
  • A) n
  • B) n²/Z
  • C) Z/n²
  • D) 1/n

Answer Key

• Q1: A - Bohr radius for hydrogen ground state is 0.53 Å.
• Q2: A - Energy at n=4 is -13.6/4² = -0.85 eV.
• Q3: B - Radius varies as n²/Z.

Memory Aids

• Radius r_n = 0.53 n²/Z Å
• Energy E_n = -13.6/n² eV
• Ground state energy = -13.6 eV

Formulas and Energy Levels

• Centripetal force: \( \frac{mv_n^2}{r_n} = \frac{1}{4\pi\epsilon_0} \frac{eZe}{r_n^2} \)
• Angular momentum: \( L_n = m v_n r_n = n \frac{h}{2\pi} \)
• Radius of nth orbit: \( r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z} = 0.53 \frac{n^2}{Z} \text{ Å} \)
• Energy levels: \( E_n = -\frac{13.6}{n^2} \text{ eV} \) for hydrogen atom (Z=1)
• Ground state (n=1) energy: \( -13.6 \text{ eV} \), higher states have less negative energy.

Solved Examples

• Ex 1: Calculate radius for \( n=2 \) in hydrogen: \( r_2 = 0.53 \times \frac{2^2}{1} = 2.12 \text{ Å} \).
• Ex 2: Energy at \( n=3 \): \( E_3 = -\frac{13.6}{3^2} = -1.51 \text{ eV} \).

Practice Set

• Q1: Radius of \( n=1 \) orbit in hydrogen is:
  • A) 0.53 Å
  • B) 1.06 Å
  • C) 2.12 Å
  • D) 0.26 Å
• Q2: Energy of electron in \( n=4 \) orbit is:
  • A) -0.85 eV
  • B) -3.4 eV
  • C) -13.6 eV
  • D) -1.51 eV
• Q3: Radius \( r_n \) is proportional to:
  • A) \( n \)
  • B) \( n^2/Z \)
  • C) \( Z/n^2 \)
  • D) \( 1/n \)

Answer Key

• Q1: A - Bohr radius for hydrogen ground state is 0.53 Å.
• Q2: A - Energy at \( n=4 \) is \( -13.6/4^2 = -0.85 \text{ eV} \).
• Q3: B - Radius varies as \( n^2/Z \).

Memory Aids

• Radius \( r_n = 0.53 \frac{n^2}{Z} \) Å
• Energy \( E_n = -\frac{13.6}{n^2} \) eV
• Ground state energy = -13.6 eV

Energy Quantization

Concept of Energy Quantization

• Electron energy in orbits is discrete, not continuous.
• Only certain energy values allowed for electron in atom.
• Explains stability of atoms and spectral lines.

Solved Examples

• Ex 1: Electron cannot have energy between -13.6 eV and -3.4 eV in hydrogen atom.

Practice Set

• Q1: Energy levels in atom are:
  • A) Continuous
  • B) Discrete
  • C) Random
  • D) Infinite
• Q2: Energy quantization explains:
  • A) Atomic stability
  • B) Continuous spectra
  • C) Electron emission only
  • D) Nuclear reactions
• Q3: Electron energy between allowed levels is:
  • A) Allowed
  • B) Forbidden
  • C) Random
  • D) Infinite

Answer Key

• Q1: B - Energy levels are discrete, not continuous.
• Q2: A - Quantization explains atomic stability and spectra.
• Q3: B - Electron cannot have energy between allowed levels.

Memory Aids

• Energy levels = discrete values only
• Electron jumps cause photon emission/absorption
• Quantization explains atomic spectra

Concept of Energy Quantization

• Electron energy in orbits is discrete, not continuous.
• Only certain energy values allowed for electron in atom.
• Explains stability of atoms and spectral lines.

Solved Examples

• Ex 1: Electron cannot have energy between \( -13.6 \text{ eV} \) and \( -3.4 \text{ eV} \) in hydrogen atom.

Practice Set

• Q1: Energy levels in atom are:
  • A) Continuous
  • B) Discrete
  • C) Random
  • D) Infinite
• Q2: Energy quantization explains:
  • A) Atomic stability
  • B) Continuous spectra
  • C) Electron emission only
  • D) Nuclear reactions
• Q3: Electron energy between allowed levels is:
  • A) Allowed
  • B) Forbidden
  • C) Random
  • D) Infinite

Answer Key

• Q1: B - Energy levels are discrete, not continuous.
• Q2: A - Quantization explains atomic stability and spectra.
• Q3: B - Electron cannot have energy between allowed levels.

Memory Aids

• Energy levels = discrete values only
• Electron jumps cause photon emission/absorption
• Quantization explains atomic spectra

Line Spectra of Hydrogen

Energy Transitions and Rydberg Formula

• Electron transitions from higher energy level n_i to lower n_f emit photon of frequency ν_if.
• Energy difference: hν_if = E_ni - E_nf = me⁴/(8ε₀²h²) (1/n_f² - 1/n_i²).
• Frequency formula: ν_if = me⁴/(8ε₀²h³) (1/n_f² - 1/n_i²).
• Discrete frequencies arise due to integer quantum numbers n_i, n_f.
• Series: Lyman (n_f=1), Balmer (n_f=2), Paschen (n_f=3), Brackett (n_f=4), Pfund (n_f=5).

Solved Examples

• Ex 1: Calculate frequency for transition n_i=3 to n_f=2 in hydrogen.
  • Use ν = me⁴/(8ε₀²h³) (1/2² - 1/3²).
  • Substitute constants and compute ν.

Practice Set

• Q1: Which series corresponds to transitions ending at n=2?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Brackett
• Q2: The frequency of emitted photon for transition n_i=4 to n_f=1 is proportional to?
  • A. 1/1² - 1/4²
  • B. 1/4² - 1/1²
  • C. 1/2² - 1/3²
  • D. 1/3² - 1/2²
• Q3: For hydrogen, which series involves transitions to n=3?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Pfund

Answer Key

• Q1: B. Balmer - Transitions ending at n=2 form Balmer series.
• Q2: A. 1/1² - 1/4² - Frequency depends on difference 1/n_f² - 1/n_i².
• Q3: C. Paschen - Transitions to n=3 define Paschen series.

Memory Aids

• Rydberg formula key: ν = R (1/n_f² - 1/n_i²).
• Series order: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).
• Higher n_i to lower n_f emits photon.

Energy Transitions and Rydberg Formula

• Electron transitions from higher energy level n_i to lower n_f emit photon of frequency ν_if.
• Energy difference: hν_if = E_ni - E_nf = me⁴/(8ε₀²h²) (1/n_f² - 1/n_i²).
• Frequency formula: ν_if = me⁴/(8ε₀²h³) (1/n_f² - 1/n_i²).
• Discrete frequencies arise due to integer quantum numbers n_i, n_f.
• Series: Lyman (n_f=1), Balmer (n_f=2), Paschen (n_f=3), Brackett (n_f=4), Pfund (n_f=5).

Solved Examples

• Ex 1: Calculate frequency for transition n_i=3 to n_f=2 in hydrogen.
  • Use ν = me⁴/(8ε₀²h³) (1/2² - 1/3²).
  • Substitute constants and compute ν.

Practice Set

• Q1: Which series corresponds to transitions ending at n=2?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Brackett
• Q2: The frequency of emitted photon for transition n_i=4 to n_f=1 is proportional to?
  • A. 1/1² - 1/4²
  • B. 1/4² - 1/1²
  • C. 1/2² - 1/3²
  • D. 1/3² - 1/2²
• Q3: For hydrogen, which series involves transitions to n=3?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Pfund

Answer Key

• Q1: B. Balmer - Transitions ending at n=2 form Balmer series.
• Q2: A. 1/1² - 1/4² - Frequency depends on difference 1/n_f² - 1/n_i².
• Q3: C. Paschen - Transitions to n=3 define Paschen series.

Memory Aids

• Rydberg formula key: ν = R (1/n_f² - 1/n_i²).
• Series order: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).
• Higher n_i to lower n_f emits photon.

Energy Transitions and Rydberg Formula

• Electron transitions from higher energy level n_i to lower n_f emit photon of frequency ν_if.
• Energy difference: hν_if = E_ni - E_nf = me⁴/(8ε₀²h²) (1/n_f² - 1/n_i²).
• Frequency formula: ν_if = me⁴/(8ε₀²h³) (1/n_f² - 1/n_i²).
• Discrete frequencies arise due to integer quantum numbers n_i, n_f.
• Series: Lyman (n_f=1), Balmer (n_f=2), Paschen (n_f=3), Brackett (n_f=4), Pfund (n_f=5).

Solved Examples

• Ex 1: Calculate frequency for transition n_i=3 to n_f=2 in hydrogen.
  • Use ν = me⁴/(8ε₀²h³) (1/2² - 1/3²).
  • Substitute constants and compute ν.

Practice Set

• Q1: Which series corresponds to transitions ending at n=2?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Brackett
• Q2: The frequency of emitted photon for transition n_i=4 to n_f=1 is proportional to?
  • A. 1/1² - 1/4²
  • B. 1/4² - 1/1²
  • C. 1/2² - 1/3²
  • D. 1/3² - 1/2²
• Q3: For hydrogen, which series involves transitions to n=3?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Pfund

Answer Key

• Q1: B. Balmer - Transitions ending at n=2 form Balmer series.
• Q2: A. 1/1² - 1/4² - Frequency depends on difference 1/n_f² - 1/n_i².
• Q3: C. Paschen - Transitions to n=3 define Paschen series.

Memory Aids

• Rydberg formula key: ν = R (1/n_f² - 1/n_i²).
• Series order: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).
• Higher n_i to lower n_f emits photon.

Energy Transitions and Rydberg Formula

• Electron transitions from higher energy level \( n_i \) to lower \( n_f \) emit photon of frequency \( \nu_{if} \).
• Energy difference: \( h\nu_{if} = E_{n_i} - E_{n_f} = \frac{me^4}{8\varepsilon_0^2 h^2} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Frequency formula: \( \nu_{if} = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Discrete frequencies arise due to integer quantum numbers \( n_i, n_f \).
• Series: Lyman (\( n_f=1 \)), Balmer (\( n_f=2 \)), Paschen (\( n_f=3 \)), Brackett (\( n_f=4 \)), Pfund (\( n_f=5 \)).

Solved Examples

• Ex 1: Calculate frequency for transition \( n_i=3 \) to \( n_f=2 \) in hydrogen.
  • Use \( \nu = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \).
  • Substitute constants and compute \( \nu \).

Practice Set

• Q1: Which series corresponds to transitions ending at \( n=2 \)?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Brackett
• Q2: The frequency of emitted photon for transition \( n_i=4 \) to \( n_f=1 \) is proportional to?
  • A. \( \frac{1}{1^2} - \frac{1}{4^2} \)
  • B. \( \frac{1}{4^2} - \frac{1}{1^2} \)
  • C. \( \frac{1}{2^2} - \frac{1}{3^2} \)
  • D. \( \frac{1}{3^2} - \frac{1}{2^2} \)
• Q3: For hydrogen, which series involves transitions to \( n=3 \)?
  • A. Lyman
  • B. Balmer
  • C. Paschen
  • D. Pfund

Answer Key

• Q1: B. Balmer - Transitions ending at \( n=2 \) form Balmer series.
• Q2: A. \( \frac{1}{1^2} - \frac{1}{4^2} \) - Frequency depends on difference \( \frac{1}{n_f^2} - \frac{1}{n_i^2} \).
• Q3: C. Paschen - Transitions to \( n=3 \) define Paschen series.

Memory Aids

• Rydberg formula key: \( \nu = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Series order: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).
• Higher \( n_i \) to lower \( n_f \) emits photon.

Continuous and Characteristic X-rays

X-ray Production and Cutoff Wavelength

• Electrons accelerated by potential V strike metal target producing X-rays.
• Cutoff wavelength: λ_min = hc/eV or λ_min (nm) = 1242/V (volt).
• λ_min depends only on accelerating voltage, not target material.

Characteristic X-rays and Electron Transitions

• Incident electron knocks out inner shell electron (K-shell), creating vacancy.
• Electron from higher shell fills vacancy, emitting characteristic X-ray photon.
• Common X-ray lines: K_α, K_β, L_α.
• Wavelength formulas:
  • λ = hc/(E_K - E_L) for K_α.
  • λ = hc/(E_K - E_M) for K_β.
  • λ = hc/(E_L - E_M) for L_α.

Solved Examples

• Ex 1: Calculate cutoff wavelength for V=2000 V.
  • Use λ_min = 1242/V = 1242/2000 = 0.621 nm.

Practice Set

• Q1: Cutoff wavelength λ_min depends on?
  • A. Target material
  • B. Accelerating voltage
  • C. Electron mass
  • D. Atomic number
• Q2: Characteristic X-rays are produced when?
  • A. Electron slows down in target
  • B. Electron knocks out inner shell electron
  • C. Electron accelerates in vacuum
  • D. Electron collides with outer shell electron
• Q3: Wavelength of K_α X-ray is given by?
  • A. hc/(E_K - E_L)
  • B. hc/(E_L - E_M)
  • C. hc/(E_K - E_M)
  • D. hc/(E_M - E_N)

Answer Key

• Q1: B. Accelerating voltage - λ_min depends only on V.
• Q2: B. Electron knocks out inner shell electron - characteristic X-rays arise from inner shell vacancies.
• Q3: A. hc/(E_K - E_L) - Formula for K_α wavelength.

Memory Aids

• λ_min = 1242/V nm for cutoff wavelength.
• Characteristic X-rays: electron transitions to inner shell vacancies.
• K, L, M shells correspond to X-ray lines K_α, K_β, L_α.

X-ray Production and Cutoff Wavelength

• Electrons accelerated by potential V strike metal target producing X-rays.
• Cutoff wavelength: λ_min = hc/eV or λ_min (nm) = 1242/V (volt).
• λ_min depends only on accelerating voltage, not target material.

Characteristic X-rays and Electron Transitions

• Incident electron knocks out inner shell electron (K-shell), creating vacancy.
• Electron from higher shell fills vacancy, emitting characteristic X-ray photon.
• Common X-ray lines: K_α, K_β, L_α.
• Wavelength formulas:
  • λ = hc/(E_K - E_L) for K_α.
  • λ = hc/(E_K - E_M) for K_β.
  • λ = hc/(E_L - E_M) for L_α.

Solved Examples

• Ex 1: Calculate cutoff wavelength for V=2000 V.
  • Use λ_min = 1242/V = 1242/2000 = 0.621 nm.

Practice Set

• Q1: Cutoff wavelength λ_min depends on?
  • A. Target material
  • B. Accelerating voltage
  • C. Electron mass
  • D. Atomic number
• Q2: Characteristic X-rays are produced when?
  • A. Electron slows down in target
  • B. Electron knocks out inner shell electron
  • C. Electron accelerates in vacuum
  • D. Electron collides with outer shell electron
• Q3: Wavelength of K_α X-ray is given by?
  • A. hc/(E_K - E_L)
  • B. hc/(E_L - E_M)
  • C. hc/(E_K - E_M)
  • D. hc/(E_M - E_N)

Answer Key

• Q1: B. Accelerating voltage - λ_min depends only on V.
• Q2: B. Electron knocks out inner shell electron - characteristic X-rays arise from inner shell vacancies.
• Q3: A. hc/(E_K - E_L) - Formula for K_α wavelength.

Memory Aids

• λ_min = 1242/V nm for cutoff wavelength.
• Characteristic X-rays: electron transitions to inner shell vacancies.
• K, L, M shells correspond to X-ray lines K_α, K_β, L_α.

X-ray Production and Cutoff Wavelength

• Electrons accelerated by potential V strike metal target producing X-rays.
• Cutoff wavelength: λ_min = hc/eV or λ_min (nm) = 1242/V (volt).
• λ_min depends only on accelerating voltage, not target material.

Characteristic X-rays and Electron Transitions

• Incident electron knocks out inner shell electron (K-shell), creating vacancy.
• Electron from higher shell fills vacancy, emitting characteristic X-ray photon.
• Common X-ray lines: K_α, K_β, L_α.
• Wavelength formulas:
  • λ = hc/(E_K - E_L) for K_α.
  • λ = hc/(E_K - E_M) for K_β.
  • λ = hc/(E_L - E_M) for L_α.

Solved Examples

• Ex 1: Calculate cutoff wavelength for V=2000 V.
  • Use λ_min = 1242/V = 1242/2000 = 0.621 nm.

Practice Set

• Q1: Cutoff wavelength λ_min depends on?
  • A. Target material
  • B. Accelerating voltage
  • C. Electron mass
  • D. Atomic number
• Q2: Characteristic X-rays are produced when?
  • A. Electron slows down in target
  • B. Electron knocks out inner shell electron
  • C. Electron accelerates in vacuum
  • D. Electron collides with outer shell electron
• Q3: Wavelength of K_α X-ray is given by?
  • A. hc/(E_K - E_L)
  • B. hc/(E_L - E_M)
  • C. hc/(E_K - E_M)
  • D. hc/(E_M - E_N)

Answer Key

• Q1: B. Accelerating voltage - λ_min depends only on V.
• Q2: B. Electron knocks out inner shell electron - characteristic X-rays arise from inner shell vacancies.
• Q3: A. hc/(E_K - E_L) - Formula for K_α wavelength.

Memory Aids

• λ_min = 1242/V nm for cutoff wavelength.
• Characteristic X-rays: electron transitions to inner shell vacancies.
• K, L, M shells correspond to X-ray lines K_α, K_β, L_α.

X-ray Production and Cutoff Wavelength

• Electrons accelerated by potential \( V \) strike metal target producing X-rays.
• Cutoff wavelength: \( \lambda_{min} = \frac{hc}{eV} \) or \( \lambda_{min} (nm) = \frac{1242}{V (volt)} \).
• \( \lambda_{min} \) depends only on accelerating voltage, not target material.

Characteristic X-rays and Electron Transitions

• Incident electron knocks out inner shell electron (K-shell), creating vacancy.
• Electron from higher shell fills vacancy, emitting characteristic X-ray photon.
• Common X-ray lines: \( K_\alpha, K_\beta, L_\alpha \).
• Wavelength formulas:
  • \( \lambda = \frac{hc}{E_K - E_L} \) for \( K_\alpha \).
  • \( \lambda = \frac{hc}{E_K - E_M} \) for \( K_\beta \).
  • \( \lambda = \frac{hc}{E_L - E_M} \) for \( L_\alpha \).

Solved Examples

• Ex 1: Calculate cutoff wavelength for \( V=2000 \) V.
  • Use \( \lambda_{min} = \frac{1242}{V} = \frac{1242}{2000} = 0.621 \) nm.

Practice Set

• Q1: Cutoff wavelength \( \lambda_{min} \) depends on?
  • A. Target material
  • B. Accelerating voltage
  • C. Electron mass
  • D. Atomic number
• Q2: Characteristic X-rays are produced when?
  • A. Electron slows down in target
  • B. Electron knocks out inner shell electron
  • C. Electron accelerates in vacuum
  • D. Electron collides with outer shell electron
• Q3: Wavelength of \( K_\alpha \) X-ray is given by?
  • A. \( \frac{hc}{E_K - E_L} \)
  • B. \( \frac{hc}{E_L - E_M} \)
  • C. \( \frac{hc}{E_K - E_M} \)
  • D. \( \frac{hc}{E_M - E_N} \)

Answer Key

• Q1: B. Accelerating voltage - \( \lambda_{min} \) depends only on \( V \).
• Q2: B. Electron knocks out inner shell electron - characteristic X-rays arise from inner shell vacancies.
• Q3: A. \( \frac{hc}{E_K - E_L} \) - Formula for \( K_\alpha \) wavelength.

Memory Aids

• \( \lambda_{min} = \frac{1242}{V} \) nm for cutoff wavelength.
• Characteristic X-rays: electron transitions to inner shell vacancies.
• K, L, M shells correspond to X-ray lines \( K_\alpha, K_\beta, L_\alpha \).

De-Broglie Explanation of Bohr’s Second Postulate

Concepts

• Bohr’s second postulate: Angular momentum quantised as L_n = n h/2π, n=1,2,3...
• De Broglie hypothesis: Electron behaves as a particle wave in circular orbit.
• Standing waves form when orbit circumference equals integral multiples of wavelength: n λ = 2π r_n
• De Broglie wavelength: λ = h/mv_n
• From n λ = 2π r_n and λ = h/mv_n, get m v_n r_n = n h/2π (Bohr’s quantisation)
• Only orbits with standing waves (resonant condition) are allowed.

Solved Examples

• Ex 1: Find angular momentum for n=2 orbit.
  • Use L_n = n h/2π
  • L₂ = 2 h/2π = h/π

Practice Set

• Q1: What is the de Broglie wavelength formula?
  • A) λ = h/mv
  • B) λ = mv
  • C) λ = mv/h
  • D) λ = h v
• Q2: For an electron in 3rd orbit, circumference is:
  • A) 2π r₃
  • B) 3 λ
  • C) Both A and B
  • D) None
• Q3: Angular momentum for n=1 orbit is:
  • A) h/2π
  • B) h
  • C) 2π h
  • D) h/π

Answer Key

• Q1: A - De Broglie wavelength is λ = h/mv.
• Q2: C - Circumference equals n λ, so both are equal.
• Q3: A - For n=1, L = h/2π.

Memory Aids

• Angular momentum quantised as multiples of h/2π.
• Standing waves form only if orbit circumference = integral × wavelength.
• De Broglie wavelength λ = h/mv.

Concepts

• Bohr’s second postulate: Angular momentum quantised as L_n = n h/2π, n=1,2,3...
• De Broglie hypothesis: Electron behaves as a particle wave in circular orbit.
• Standing waves form when orbit circumference equals integral multiples of wavelength: n λ = 2π r_n
• De Broglie wavelength: λ = h/mv_n
• From n λ = 2π r_n and λ = h/mv_n, get m v_n r_n = n h/2π (Bohr’s quantisation)
• Only orbits with standing waves (resonant condition) are allowed.

Solved Examples

• Ex 1: Find angular momentum for n=2 orbit.
  • Use L_n = n h/2π
  • L₂ = 2 h/2π = h/π

Practice Set

• Q1: What is the de Broglie wavelength formula?
  • A) λ = h/mv
  • B) λ = mv
  • C) λ = mv/h
  • D) λ = h v
• Q2: For an electron in 3rd orbit, circumference is:
  • A) 2π r₃
  • B) 3 λ
  • C) Both A and B
  • D) None
• Q3: Angular momentum for n=1 orbit is:
  • A) h/2π
  • B) h
  • C) 2π h
  • D) h/π

Answer Key

• Q1: A - De Broglie wavelength is λ = h/mv.
• Q2: C - Circumference equals n λ, so both are equal.
• Q3: A - For n=1, L = h/2π.

Memory Aids

• Angular momentum quantised as multiples of h/2π.
• Standing waves form only if orbit circumference = integral × wavelength.
• De Broglie wavelength λ = h/mv.

Concepts

• Bohr’s second postulate: Angular momentum quantised as L_n = n h/2π, n=1,2,3...
• De Broglie hypothesis: Electron behaves as a particle wave in circular orbit.
• Standing waves form when orbit circumference equals integral multiples of wavelength: n λ = 2π r_n
• De Broglie wavelength: λ = h/mv_n
• From n λ = 2π r_n and λ = h/mv_n, get m v_n r_n = n h/2π (Bohr’s quantisation)
• Only orbits with standing waves (resonant condition) are allowed.

Solved Examples

• Ex 1: Find angular momentum for n=2 orbit.
  • Use L_n = n h/2π
  • L₂ = 2 h/2π = h/π

Practice Set

• Q1: What is the de Broglie wavelength formula?
  • A) λ = h/mv
  • B) λ = mv
  • C) λ = mv/h
  • D) λ = h v
• Q2: For an electron in 3rd orbit, circumference is:
  • A) 2π r₃
  • B) 3 λ
  • C) Both A and B
  • D) None
• Q3: Angular momentum for n=1 orbit is:
  • A) h/2π
  • B) h
  • C) 2π h
  • D) h/π

Answer Key

• Q1: A - De Broglie wavelength is λ = h/mv.
• Q2: C - Circumference equals n λ, so both are equal.
• Q3: A - For n=1, L = h/2π.

Memory Aids

• Angular momentum quantised as multiples of h/2π.
• Standing waves form only if orbit circumference = integral × wavelength.
• De Broglie wavelength λ = h/mv.

Concepts

• Bohr’s second postulate: Angular momentum quantised as \( L_n = \frac{n h}{2\pi} \), \( n=1,2,3... \)
• De Broglie hypothesis: Electron behaves as a particle wave in circular orbit.
• Standing waves form when orbit circumference equals integral multiples of wavelength: \( n \lambda = 2\pi r_n \)
• De Broglie wavelength: \( \lambda = \frac{h}{mv_n} \)
• From \( n \lambda = 2\pi r_n \) and \( \lambda = \frac{h}{mv_n} \), get \( m v_n r_n = \frac{n h}{2\pi} \) (Bohr’s quantisation)
• Only orbits with standing waves (resonant condition) are allowed.

Solved Examples

• Ex 1: Find angular momentum for \( n=2 \) orbit.
  • Use \( L_n = \frac{n h}{2\pi} \)
  • \( L_2 = \frac{2 h}{2\pi} = \frac{h}{\pi} \)

Practice Set

• Q1: What is the de Broglie wavelength formula?
  • A) \( \lambda = \frac{h}{mv} \)
  • B) \( \lambda = mv \)
  • C) \( \lambda = \frac{mv}{h} \)
  • D) \( \lambda = h v \)
• Q2: For an electron in 3rd orbit, circumference is:
  • A) \( 2\pi r_3 \)
  • B) \( 3 \lambda \)
  • C) Both A and B
  • D) None
• Q3: Angular momentum for \( n=1 \) orbit is:
  • A) \( \frac{h}{2\pi} \)
  • B) \( h \)
  • C) \( 2\pi h \)
  • D) \( \frac{h}{\pi} \)

Answer Key

• Q1: A - De Broglie wavelength is \( \lambda = \frac{h}{mv} \).
• Q2: C - Circumference equals \( n \lambda \), so both are equal.
• Q3: A - For \( n=1 \), \( L = \frac{h}{2\pi} \).

Memory Aids

• Angular momentum quantised as multiples of \( \frac{h}{2\pi} \).
• Standing waves form only if orbit circumference = integral \( \times \) wavelength.
• De Broglie wavelength \( \lambda = \frac{h}{mv} \).

Limitations of Bohr’s Theory

Concepts

• Valid only for single-electron atoms (H, He⁺, Li²⁺).
• Cannot explain fine structure of spectral lines.
• Fails to explain molecular bonding and stability.
• Cannot fully explain Zeeman and Stark effects.
• Does not explain electron distribution in orbits.
• Ignores electron-electron interactions in multi-electron atoms.
• Cannot explain intensity variations in hydrogen emission spectrum.

Solved Examples

• Ex 1: Identify if Bohr’s theory applies to He atom.
  • He has 2 electrons → multi-electron atom.
  • Bohr’s theory fails here.

Practice Set

• Q1: Bohr’s theory is valid for:
  • A) Multi-electron atoms
  • B) Single-electron atoms
  • C) Molecules
  • D) None
• Q2: Which effect is NOT explained by Bohr’s theory?
  • A) Zeeman effect
  • B) Stark effect
  • C) Fine structure
  • D) All of the above
• Q3: Bohr’s theory ignores:
  • A) Electron-nucleus force
  • B) Electron-electron force
  • C) Nuclear force
  • D) Gravitational force

Answer Key

• Q1: B - Valid only for single-electron atoms.
• Q2: D - Bohr’s theory fails to explain all listed effects.
• Q3: B - Ignores electron-electron interactions.

Memory Aids

• Bohr’s model = single-electron only.
• Fails for fine structure, Zeeman, Stark effects.
• Ignores electron-electron repulsion.

Moseley’s Law

Concepts

• X-ray frequency relation: √ν = a(Z - b)
• Constants a, b; b ≈ 1
• Bohr’s model fits if Z replaced by Z - b
• Energy released for n=2 → n=1: ΔE = h ν = R h c (Z - b)² (1 - 1/4)

Solved Examples

• Ex 1: Calculate √ν for Z=30, b=1, a=constant.
  • √ν = a(Z - b) = a(30 - 1) = 29a

Practice Set

• Q1: Moseley’s law relates √ν to:
  • A) Atomic number Z
  • B) Z - b
  • C) Wavelength λ
  • D) Energy E
• Q2: In Moseley’s law, constant b is approximately:
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q3: Energy released in transition n=2 → n=1 depends on:
  • A) (Z - b)²
  • B) Z²
  • C) b²
  • D) None

Answer Key

• Q1: B - Frequency depends on effective atomic number Z - b.
• Q2: B - Constant b ≈ 1.
• Q3: A - Energy depends on square of Z - b.

Memory Aids

• √ν linear with Z - b.
• b ≈ 1 adjusts atomic number.
• Energy ∝ (Z - b)².

Concepts

• X-ray frequency relation: √ν = a(Z - b)
• Constants a, b; b ≈ 1
• Bohr’s model fits if Z replaced by Z - b
• Energy released for n=2 → n=1: ΔE = h ν = R h c (Z - b)² (1 - 1/4)

Solved Examples

• Ex 1: Calculate √ν for Z=30, b=1, a=constant.
  • √ν = a(Z - b) = a(30 - 1) = 29a

Practice Set

• Q1: Moseley’s law relates √ν to:
  • A) Atomic number Z
  • B) Z - b
  • C) Wavelength λ
  • D) Energy E
• Q2: In Moseley’s law, constant b is approximately:
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q3: Energy released in transition n=2 → n=1 depends on:
  • A) (Z - b)²
  • B) Z²
  • C) b²
  • D) None

Answer Key

• Q1: B - Frequency depends on effective atomic number Z - b.
• Q2: B - Constant b ≈ 1.
• Q3: A - Energy depends on square of Z - b.

Memory Aids

• √ν linear with Z - b.
• b ≈ 1 adjusts atomic number.
• Energy ∝ (Z - b)².

Concepts

• X-ray frequency relation: √ν = a(Z - b)
• Constants a, b; b ≈ 1
• Bohr’s model fits if Z replaced by Z - b
• Energy released for n=2 → n=1: ΔE = h ν = R h c (Z - b)² (1 - 1/4)

Solved Examples

• Ex 1: Calculate √ν for Z=30, b=1, a=constant.
  • √ν = a(Z - b) = a(30 - 1) = 29a

Practice Set

• Q1: Moseley’s law relates √ν to:
  • A) Atomic number Z
  • B) Z - b
  • C) Wavelength λ
  • D) Energy E
• Q2: In Moseley’s law, constant b is approximately:
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q3: Energy released in transition n=2 → n=1 depends on:
  • A) (Z - b)²
  • B) Z²
  • C) b²
  • D) None

Answer Key

• Q1: B - Frequency depends on effective atomic number Z - b.
• Q2: B - Constant b ≈ 1.
• Q3: A - Energy depends on square of Z - b.

Memory Aids

• √ν linear with Z - b.
• b ≈ 1 adjusts atomic number.
• Energy ∝ (Z - b)².

Concepts

• X-ray frequency relation: √ν = a(Z - b)
• Constants a, b; b ≈ 1
• Bohr’s model fits if Z replaced by Z - b
• Energy released for n=2 → n=1: ΔE = h ν = R h c (Z - b)² (1 - 1/4)

Solved Examples

• Ex 1: Calculate √ν for Z=30, b=1, a=constant.
  • √ν = a(Z - b) = a(30 - 1) = 29a

Practice Set

• Q1: Moseley’s law relates √ν to:
  • A) Atomic number Z
  • B) Z - b
  • C) Wavelength λ
  • D) Energy E
• Q2: In Moseley’s law, constant b is approximately:
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q3: Energy released in transition n=2 → n=1 depends on:
  • A) (Z - b)²
  • B) Z²
  • C) b²
  • D) None

Answer Key

• Q1: B - Frequency depends on effective atomic number Z - b.
• Q2: B - Constant b ≈ 1.
• Q3: A - Energy depends on square of Z - b.

Memory Aids

• √ν linear with Z - b.
• b ≈ 1 adjusts atomic number.
• Energy ∝ (Z - b)².

Concepts

• X-ray frequency relation: \( \sqrt{\nu} = a(Z - b) \)
• Constants \( a, b \); \( b \approx 1 \)
• Bohr’s model fits if \( Z \) replaced by \( Z - b \)
• Energy released for \( n=2 \to n=1 \): \( \Delta E = h \nu = R h c (Z - b)^2 (1 - \frac{1}{4}) \)

Solved Examples

• Ex 1: Calculate \( \sqrt{\nu} \) for \( Z=30, b=1, a=constant \).
  • \( \sqrt{\nu} = a(Z - b) = a(30 - 1) = 29a \)

Practice Set

• Q1: Moseley’s law relates \( \sqrt{\nu} \) to:
  • A) Atomic number \( Z \)
  • B) \( Z - b \)
  • C) Wavelength \( \lambda \)
  • D) Energy \( E \)
• Q2: In Moseley’s law, constant \( b \) is approximately:
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q3: Energy released in transition \( n=2 \to n=1 \) depends on:
  • A) \( (Z - b)^2 \)
  • B) \( Z^2 \)
  • C) \( b^2 \)
  • D) None

Answer Key

• Q1: B - Frequency depends on effective atomic number \( Z - b \).
• Q2: B - Constant \( b \approx 1 \).
• Q3: A - Energy depends on square of \( Z - b \).

Memory Aids

• \( \sqrt{\nu} \) linear with \( Z - b \).
• \( b \approx 1 \) adjusts atomic number.
• Energy \( \propto (Z - b)^2 \).

Bragg’s Law

Concepts

• Condition for X-ray diffraction: 2 d sin θ = n λ
• n: order of spectrum (1,2,3...)
• d: distance between crystal planes (m)
• θ: glancing angle
• λ: X-ray wavelength (m)
• Path difference between waves: d sin θ
• Constructive interference occurs when path difference = integral multiples of wavelength.

Solved Examples

• Ex 1: Calculate θ for first order diffraction with d=0.2 nm, λ=0.1 nm.
  • Use 2 d sin θ = n λ
  • 2 × 0.2 × sin θ = 1 × 0.1 → sin θ = 0.25
  • θ = sin^-1 0.25 = 14.48°

Practice Set

• Q1: Bragg’s law formula is:
  • A) 2 d sin θ = n λ
  • B) d = 2 λ sin θ
  • C) n = 2 d sin θ
  • D) λ = 2 d sin θ
• Q2: In Bragg’s law, d is:
  • A) Distance between atoms
  • B) Distance between adjacent planes
  • C) Wavelength
  • D) Glancing angle
• Q3: For n=2, d=0.3 nm, λ=0.15 nm, find θ.
  • A) 30°
  • B) 45°
  • C) 60°
  • D) 90°

Answer Key

• Q1: A - Bragg’s law is 2 d sin θ = n λ.
• Q2: B - d is distance between adjacent crystal planes.
• Q3: A - 2 × 0.3 × sin θ = 2 × 0.15 → sin θ = 0.5 → θ = 30°.

Memory Aids

• Bragg’s law: 2 d sin θ = n λ.
• d = interplanar spacing.
• Constructive interference requires path difference = integral multiples of wavelength.

Concepts

• Condition for X-ray diffraction: 2 d sin θ = n λ
• n: order of spectrum (1,2,3...)
• d: distance between crystal planes (m)
• θ: glancing angle
• λ: X-ray wavelength (m)
• Path difference between waves: d sin θ
• Constructive interference occurs when path difference = integral multiples of wavelength.

Solved Examples

• Ex 1: Calculate θ for first order diffraction with d=0.2 nm, λ=0.1 nm.
  • Use 2 d sin θ = n λ
  • 2 × 0.2 × sin θ = 1 × 0.1 → sin θ = 0.25
  • θ = sin^-1 0.25 = 14.48°

Practice Set

• Q1: Bragg’s law formula is:
  • A) 2 d sin θ = n λ
  • B) d = 2 λ sin θ
  • C) n = 2 d sin θ
  • D) λ = 2 d sin θ
• Q2: In Bragg’s law, d is:
  • A) Distance between atoms
  • B) Distance between adjacent planes
  • C) Wavelength
  • D) Glancing angle
• Q3: For n=2, d=0.3 nm, λ=0.15 nm, find θ.
  • A) 30°
  • B) 45°
  • C) 60°
  • D) 90°

Answer Key

• Q1: A - Bragg’s law is 2 d sin θ = n λ.
• Q2: B - d is distance between adjacent crystal planes.
• Q3: A - 2 × 0.3 × sin θ = 2 × 0.15 → sin θ = 0.5 → θ = 30°.

Memory Aids

• Bragg’s law: 2 d sin θ = n λ.
• d = interplanar spacing.
• Constructive interference requires path difference = integral multiples of wavelength.

Concepts

• Condition for X-ray diffraction: 2 d sin θ = n λ
• n: order of spectrum (1,2,3...)
• d: distance between crystal planes (m)
• θ: glancing angle
• λ: X-ray wavelength (m)
• Path difference between waves: d sin θ
• Constructive interference occurs when path difference = integral multiples of wavelength.

Solved Examples

• Ex 1: Calculate θ for first order diffraction with d=0.2 nm, λ=0.1 nm.
  • Use 2 d sin θ = n λ
  • 2 × 0.2 × sin θ = 1 × 0.1 → sin θ = 0.25
  • θ = sin^-1 0.25 = 14.48°

Practice Set

• Q1: Bragg’s law formula is:
  • A) 2 d sin θ = n λ
  • B) d = 2 λ sin θ
  • C) n = 2 d sin θ
  • D) λ = 2 d sin θ
• Q2: In Bragg’s law, d is:
  • A) Distance between atoms
  • B) Distance between adjacent planes
  • C) Wavelength
  • D) Glancing angle
• Q3: For n=2, d=0.3 nm, λ=0.15 nm, find θ.
  • A) 30°
  • B) 45°
  • C) 60°
  • D) 90°

Answer Key

• Q1: A - Bragg’s law is 2 d sin θ = n λ.
• Q2: B - d is distance between adjacent crystal planes.
• Q3: A - 2 × 0.3 × sin θ = 2 × 0.15 → sin θ = 0.5 → θ = 30°.

Memory Aids

• Bragg’s law: 2 d sin θ = n λ.
• d = interplanar spacing.
• Constructive interference requires path difference = integral multiples of wavelength.

Concepts

• Condition for X-ray diffraction: 2 d sin θ = n λ
• n: order of spectrum (1,2,3...)
• d: distance between crystal planes (m)
• θ: glancing angle
• λ: X-ray wavelength (m)
• Path difference between waves: d sin θ
• Constructive interference occurs when path difference = integral multiples of wavelength.

Solved Examples

• Ex 1: Calculate θ for first order diffraction with d=0.2 nm, λ=0.1 nm.
  • Use 2 d sin θ = n λ
  • 2 × 0.2 × sin θ = 1 × 0.1 → sin θ = 0.25
  • θ = sin^-1 0.25 = 14.48°

Practice Set

• Q1: Bragg’s law formula is:
  • A) 2 d sin θ = n λ
  • B) d = 2 λ sin θ
  • C) n = 2 d sin θ
  • D) λ = 2 d sin θ
• Q2: In Bragg’s law, d is:
  • A) Distance between atoms
  • B) Distance between adjacent planes
  • C) Wavelength
  • D) Glancing angle
• Q3: For n=2, d=0.3 nm, λ=0.15 nm, find θ.
  • A) 30°
  • B) 45°
  • C) 60°
  • D) 90°

Answer Key

• Q1: A - Bragg’s law is 2 d sin θ = n λ.
• Q2: B - d is distance between adjacent crystal planes.
• Q3: A - 2 × 0.3 × sin θ = 2 × 0.15 → sin θ = 0.5 → θ = 30°.

Memory Aids

• Bragg’s law: 2 d sin θ = n λ.
• d = interplanar spacing.
• Constructive interference requires path difference = integral multiples of wavelength.

Concepts

• Condition for X-ray diffraction: \( 2 d \sin \theta = n \lambda \)
• \( n \): order of spectrum (1,2,3...)
• \( d \): distance between crystal planes (m)
• \( \theta \): glancing angle
• \( \lambda \): X-ray wavelength (m)
• Path difference between waves: \( d \sin \theta \)
• Constructive interference occurs when path difference = integral multiples of wavelength.

Solved Examples

• Ex 1: Calculate \( \theta \) for first order diffraction with \( d=0.2 \text{ nm} \), \( \lambda=0.1 \text{ nm} \).
  • Use \( 2 d \sin \theta = n \lambda \)
  • \( 2 \times 0.2 \times \sin \theta = 1 \times 0.1 \Rightarrow \sin \theta = 0.25 \)
  • \( \theta = \sin^{-1} 0.25 = 14.48^\circ \)

Practice Set

• Q1: Bragg’s law formula is:
  • A) \( 2 d \sin \theta = n \lambda \)
  • B) \( d = 2 \lambda \sin \theta \)
  • C) \( n = 2 d \sin \theta \)
  • D) \( \lambda = 2 d \sin \theta \)
• Q2: In Bragg’s law, \( d \) is:
  • A) Distance between atoms
  • B) Distance between adjacent planes
  • C) Wavelength
  • D) Glancing angle
• Q3: For \( n=2 \), \( d=0.3 \text{ nm} \), \( \lambda=0.15 \text{ nm} \), find \( \theta \).
  • A) 30°
  • B) 45°
  • C) 60°
  • D) 90°

Answer Key

• Q1: A - Bragg’s law is \( 2 d \sin \theta = n \lambda \).
• Q2: B - \( d \) is distance between adjacent crystal planes.
• Q3: A - \( 2 \times 0.3 \times \sin \theta = 2 \times 0.15 \Rightarrow \sin \theta = 0.5 \Rightarrow \theta = 30^\circ \).

Memory Aids

• Bragg’s law: \( 2 d \sin \theta = n \lambda \).
• \( d \) = interplanar spacing.
• Constructive interference requires path difference = integral multiples of wavelength.