Dual Nature of Radiation and Matter

NEET Ug . NEET Physics . Physics . Dual Nature of Radiation and Matter

Chapter 26: Dual Nature of Radiation and Matter

Ch 26

NEET Physics

NEET Ug

Introduction

Historical Discoveries

Maxwell’s equations and Hertz’s experiments (1887) confirmed light's wave nature.
Electric discharge in low-pressure gases led to key atomic discoveries.
X-rays discovered by Roentgen (1895); electron discovered by J.J. Thomson (1897).
Discharge tube at ~0.001 mm Hg shows glow due to cathode rays.
Glow color depends on glass type (e.g., yellowish-green for soda glass).
Cathode rays: streams of fast-moving negatively charged particles (William Crookes, 1870).
J.J. Thomson measured speed (0.1–0.2 c) and specific charge (e/m = 1.76 × 10¹¹ C/kg) of cathode rays.
Electron identified as universal particle, named by Thomson in 1897.
Millikan’s oil-drop experiment (1913) measured electron charge: 1.602 × 10^-19 C; established charge quantization.

Solved Examples

Ex 1: Calculate speed of cathode ray if it is 0.15 times speed of light.
- Speed = 0.15 × 3 × 10⁸ m/s = 4.5 × 10⁷ m/s
Ex 2: Find mass of electron given e = 1.602 × 10^-19 C and e/m = 1.76 × 10¹¹ C/kg.
- Mass, m = e / (e/m) = (1.602 × 10^-19) / (1.76 × 10¹¹) = 9.1 × 10^-31 kg

Practice Set

Q1: Who discovered the electron?
- A. Roentgen
- B. William Crookes
- C. J.J. Thomson
- D. R.A. Millikan
Q2: The specific charge (e/m) of electron is approximately:
- A. 1.76 × 10¹¹ C/kg
- B. 9.1 × 10^-31 C/kg
- C. 1.602 × 10^-19 C/kg
- D. 3 × 10⁸ C/kg
Q3: Millikan’s oil-drop experiment established:
- A. Electron mass
- B. Charge quantization
- C. Speed of electron
- D. Work function

Answer Key

Q1: C - J.J. Thomson discovered the electron in 1897.
Q2: A - Specific charge e/m = 1.76 × 10¹¹ C/kg is a known constant for electrons.
Q3: B - Millikan’s experiment proved electric charge is quantized.

Memory Aids

Electron speed ~ 0.1–0.2 × speed of light.
e/m = 1.76 × 10¹¹ C/kg (constant for electrons).
Charge of electron = 1.602 × 10^-19 C (elementary charge).

Electron Emission

Concept and Work Function

Metals have free electrons responsible for conductivity.
Electrons are held inside metal by positive ion attraction.
Electron emission: electrons escape metal surface if energy > work function.
Work function: minimum energy to free electron from metal, measured in eV.
1 eV = 1.602 × 10^-19 J (energy gained by electron accelerated through 1 V).

Types of Electron Emission

Thermionic emission: heating metal supplies thermal energy to free electrons.
Field emission: strong electric field (~10⁸ V/m) pulls electrons out (e.g., spark plug).
Photoelectric emission: light of suitable frequency causes electron emission (photoelectrons).

Solved Examples

Ex 1: Calculate energy in joules for 2 eV.
- Energy = 2 × 1.602 × 10^-19 J = 3.204 × 10^-19 J
Ex 2: Identify emission type when electrons are emitted by heating metal.
- Answer: Thermionic emission

Practice Set

Q1: What is the unit of work function?
- A. Volt
- B. Electron volt (eV)
- C. Joule
- D. Coulomb
Q2: Which emission requires strong electric field?
- A. Thermionic
- B. Field
- C. Photoelectric
- D. None
Q3: Photoelectrons are emitted due to:
- A. Heating
- B. Electric field
- C. Light illumination
- D. Magnetic field

Answer Key

Q1: B - Work function is measured in electron volts (eV).
Q2: B - Field emission requires very strong electric field (~10⁸ V/m).
Q3: C - Photoelectric emission occurs when light shines on metal.

Memory Aids

Work function energy = minimum energy to free electron.
1 eV = 1.602 × 10^-19 J.
Thermionic = heat, Field = strong electric field, Photoelectric = light.

Photoelectric Effect

Key Observations

Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet light.
Hallwachs and Lenard found that ultraviolet light on a metal plate causes current flow due to electron emission.
Photoelectric emission depends on frequency and intensity of incident light and collector plate potential.
Electrons emitted are called photoelectrons.
No electrons are emitted if the light frequency is below a threshold frequency ν₀.
Metals like zinc, cadmium, magnesium emit electrons under UV light; alkali metals emit even under visible light.

Solved Examples

Ex 1: Calculate the threshold frequency if the work function φ of a metal is 2 eV.
- Use φ = hν₀
- ν₀ = φ/h = (2 × 1.6 × 10^-19)/(6.63 × 10^-34) = 4.83 × 10¹⁴ Hz

Practice Set

Q1: What happens when ultraviolet light falls on a metal surface?
- A) Metal heats up
- B) Electrons are emitted
- C) Metal changes color
- D) No effect
Q2: The minimum frequency of light required to emit electrons is called:
- A) Cutoff frequency
- B) Threshold frequency
- C) Resonance frequency
- D) Natural frequency
Q3: If the work function of a metal is 3 eV, what is the threshold frequency? (Use h = 6.63 × 10^-34 Js, 1 eV = 1.6 × 10^-19 J)
- A) 7.2 × 10¹⁴ Hz
- B) 4.5 × 10¹⁴ Hz
- C) 1.2 × 10¹⁵ Hz
- D) 3.0 × 10¹⁴ Hz

Answer Key

Q1: B - Ultraviolet light causes emission of electrons (photoelectric effect).
Q2: B - Threshold frequency is the minimum frequency to emit electrons.
Q3: A - ν₀ = φ/h = (3 × 1.6 × 10^-19)/(6.63 × 10^-34) = 7.2 × 10¹⁴ Hz.

Memory Aids

Photoelectric effect: Light → electrons emitted.
Threshold frequency ν₀ = φ/h.
Only light with ν ≥ ν₀ causes emission.
Alkali metals emit electrons even with visible light.

Key Observations

Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet light.
Hallwachs and Lenard found that ultraviolet light on a metal plate causes current flow due to electron emission.
Photoelectric emission depends on frequency and intensity of incident light and collector plate potential.
Electrons emitted are called photoelectrons.
No electrons are emitted if the light frequency is below a threshold frequency ν₀.
Metals like zinc, cadmium, magnesium emit electrons under UV light; alkali metals emit even under visible light.

Solved Examples

Ex 1: Calculate the threshold frequency if the work function φ of a metal is 2 eV.
- Use φ = hν₀
- ν₀ = φ/h = (2 × 1.6 × 10^-19)/(6.63 × 10^-34) = 4.83 × 10¹⁴ Hz

Practice Set

Q1: What happens when ultraviolet light falls on a metal surface?
- A) Metal heats up
- B) Electrons are emitted
- C) Metal changes color
- D) No effect
Q2: The minimum frequency of light required to emit electrons is called:
- A) Cutoff frequency
- B) Threshold frequency
- C) Resonance frequency
- D) Natural frequency
Q3: If the work function of a metal is 3 eV, what is the threshold frequency? (Use h = 6.63 × 10^-34 Js, 1 eV = 1.6 × 10^-19 J)
- A) 7.2 × 10¹⁴ Hz
- B) 4.5 × 10¹⁴ Hz
- C) 1.2 × 10¹⁵ Hz
- D) 3.0 × 10¹⁴ Hz

Answer Key

Q1: B - Ultraviolet light causes emission of electrons (photoelectric effect).
Q2: B - Threshold frequency is the minimum frequency to emit electrons.
Q3: A - ν₀ = φ/h = (3 × 1.6 × 10^-19)/(6.63 × 10^-34) = 7.2 × 10¹⁴ Hz.

Memory Aids

Photoelectric effect: Light → electrons emitted.
Threshold frequency ν₀ = φ/h.
Only light with ν ≥ ν₀ causes emission.
Alkali metals emit electrons even with visible light.

Key Observations

Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet light.
Hallwachs and Lenard found that ultraviolet light on a metal plate causes current flow due to electron emission.
Photoelectric emission depends on frequency and intensity of incident light and collector plate potential.
Electrons emitted are called photoelectrons.
No electrons are emitted if the light frequency is below a threshold frequency ν₀.
Metals like zinc, cadmium, magnesium emit electrons under UV light; alkali metals emit even under visible light.

Solved Examples

Ex 1: Calculate the threshold frequency if the work function φ of a metal is 2 eV.
- Use φ = hν₀
- ν₀ = φ/h = (2 × 1.6 × 10^-19)/(6.63 × 10^-34) = 4.83 × 10¹⁴ Hz

Practice Set

Q1: What happens when ultraviolet light falls on a metal surface?
- A) Metal heats up
- B) Electrons are emitted
- C) Metal changes color
- D) No effect
Q2: The minimum frequency of light required to emit electrons is called:
- A) Cutoff frequency
- B) Threshold frequency
- C) Resonance frequency
- D) Natural frequency
Q3: If the work function of a metal is 3 eV, what is the threshold frequency? (Use h = 6.63 × 10^-34 Js, 1 eV = 1.6 × 10^-19 J)
- A) 7.2 × 10¹⁴ Hz
- B) 4.5 × 10¹⁴ Hz
- C) 1.2 × 10¹⁵ Hz
- D) 3.0 × 10¹⁴ Hz

Answer Key

Q1: B - Ultraviolet light causes emission of electrons (photoelectric effect).
Q2: B - Threshold frequency is the minimum frequency to emit electrons.
Q3: A - ν₀ = φ/h = (3 × 1.6 × 10^-19)/(6.63 × 10^-34) = 7.2 × 10¹⁴ Hz.

Memory Aids

Photoelectric effect: Light → electrons emitted.
Threshold frequency ν₀ = φ/h.
Only light with ν ≥ ν₀ causes emission.
Alkali metals emit electrons even with visible light.

Key Observations

Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet light.
Hallwachs and Lenard found that ultraviolet light on a metal plate causes current flow due to electron emission.
Photoelectric emission depends on frequency and intensity of incident light and collector plate potential.
Electrons emitted are called photoelectrons.
No electrons are emitted if the light frequency is below a threshold frequency ν₀.
Metals like zinc, cadmium, magnesium emit electrons under UV light; alkali metals emit even under visible light.

Solved Examples

Ex 1: Calculate the threshold frequency if the work function φ of a metal is 2 eV.
- Use φ = hν₀
- ν₀ = φ/h = (2 × 1.6 × 10^-19)/(6.63 × 10^-34) = 4.83 × 10¹⁴ Hz

Practice Set

Q1: What happens when ultraviolet light falls on a metal surface?
- A) Metal heats up
- B) Electrons are emitted
- C) Metal changes color
- D) No effect
Q2: The minimum frequency of light required to emit electrons is called:
- A) Cutoff frequency
- B) Threshold frequency
- C) Resonance frequency
- D) Natural frequency
Q3: If the work function of a metal is 3 eV, what is the threshold frequency? (Use h = 6.63 × 10^-34 Js, 1 eV = 1.6 × 10^-19 J)
- A) 7.2 × 10¹⁴ Hz
- B) 4.5 × 10¹⁴ Hz
- C) 1.2 × 10¹⁵ Hz
- D) 3.0 × 10¹⁴ Hz

Answer Key

Q1: B - Ultraviolet light causes emission of electrons (photoelectric effect).
Q2: B - Threshold frequency is the minimum frequency to emit electrons.
Q3: A - ν₀ = φ/h = (3 × 1.6 × 10^-19)/(6.63 × 10^-34) = 7.2 × 10¹⁴ Hz.

Memory Aids

Photoelectric effect: Light → electrons emitted.
Threshold frequency ν₀ = φ/h.
Only light with ν ≥ ν₀ causes emission.
Alkali metals emit electrons even with visible light.

Key Observations

Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet light.
Hallwachs and Lenard found that ultraviolet light on a metal plate causes current flow due to electron emission.
Photoelectric emission depends on frequency and intensity of incident light and collector plate potential.
Electrons emitted are called photoelectrons.
No electrons are emitted if the light frequency is below a threshold frequency \( \nu_0 \).
Metals like zinc, cadmium, magnesium emit electrons under UV light; alkali metals emit even under visible light.

Solved Examples

Ex 1: Calculate the threshold frequency if the work function \( \phi \) of a metal is 2 eV.
- Use \( \phi = h \nu_0 \)
- \( \nu_0 = \frac{\phi}{h} = \frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.83 \times 10^{14} \text{ Hz} \)

Practice Set

Q1: What happens when ultraviolet light falls on a metal surface?
- A) Metal heats up
- B) Electrons are emitted
- C) Metal changes color
- D) No effect
Q2: The minimum frequency of light required to emit electrons is called:
- A) Cutoff frequency
- B) Threshold frequency
- C) Resonance frequency
- D) Natural frequency
Q3: If the work function of a metal is 3 eV, what is the threshold frequency? (Use \( h = 6.63 \times 10^{-34} \text{ Js} \), \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \))
- A) \( 7.2 \times 10^{14} \text{ Hz} \)
- B) \( 4.5 \times 10^{14} \text{ Hz} \)
- C) \( 1.2 \times 10^{15} \text{ Hz} \)
- D) \( 3.0 \times 10^{14} \text{ Hz} \)

Answer Key

Q1: B - Ultraviolet light causes emission of electrons (photoelectric effect).
Q2: B - Threshold frequency is the minimum frequency to emit electrons.
Q3: A - \( \nu_0 = \frac{\phi}{h} = \frac{3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 7.2 \times 10^{14} \text{ Hz} \).

Memory Aids

Photoelectric effect: Light \( \to \) electrons emitted.
Threshold frequency \( \nu_0 = \frac{\phi}{h} \).
Only light with \( \nu \geq \nu_0 \) causes emission.
Alkali metals emit electrons even with visible light.

Experimental Study Photoelectric Effect

Setup and Working

Emitting electrode C (photosensitive) at negative potential; collecting electrode A at positive potential in evacuated tube.
Light of frequency ν > threshold frequency ν₀ causes emission of photoelectrons from C.
Photoelectrons travel to A, creating photoelectric current I proportional to number of emitted electrons.
If A is negative w.r.t C, retarding electric field reduces photoelectron flow, decreasing current.
Stopping potential V₀: retarding potential where photoelectric current becomes zero.

Solved Examples

Ex 1: Calculate stopping potential if max kinetic energy of photoelectrons is 2 × 10^-19 J.
- Use K_max = eV₀
- V₀ = K_max/e = (2 × 10^-19)/(1.6 × 10^-19) = 1.25 V

Practice Set

Q1: What happens to photoelectric current if light intensity increases?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q2: Stopping potential is the potential at which:
- A) Photoelectric current is maximum
- B) No photoelectron reaches collector
- C) Emission starts
- D) Intensity is maximum
Q3: If stopping potential is 2 V, max kinetic energy of photoelectrons is:
- A) 3.2 × 10^-19 J
- B) 1.6 × 10^-19 J
- C) 2 × 10^-19 J
- D) 4 × 10^-19 J

Answer Key

Q1: C - Photoelectric current is directly proportional to light intensity.
Q2: B - Stopping potential is where no photoelectron reaches collector, current zero.
Q3: D - K_max = eV₀ = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J (corrected to option A).

Memory Aids

Stopping potential V₀: voltage to stop photoelectrons.
Photoelectric current ∝ light intensity.
Retarding potential reduces electron flow.

Setup and Working

Emitting electrode C (photosensitive) at negative potential; collecting electrode A at positive potential in evacuated tube.
Light of frequency ν > threshold frequency ν₀ causes emission of photoelectrons from C.
Photoelectrons travel to A, creating photoelectric current I proportional to number of emitted electrons.
If A is negative w.r.t C, retarding electric field reduces photoelectron flow, decreasing current.
Stopping potential V₀: retarding potential where photoelectric current becomes zero.

Solved Examples

Ex 1: Calculate stopping potential if max kinetic energy of photoelectrons is 2 × 10^-19 J.
- Use K_max = eV₀
- V₀ = K_max/e = (2 × 10^-19)/(1.6 × 10^-19) = 1.25 V

Practice Set

Q1: What happens to photoelectric current if light intensity increases?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q2: Stopping potential is the potential at which:
- A) Photoelectric current is maximum
- B) No photoelectron reaches collector
- C) Emission starts
- D) Intensity is maximum
Q3: If stopping potential is 2 V, max kinetic energy of photoelectrons is:
- A) 3.2 × 10^-19 J
- B) 1.6 × 10^-19 J
- C) 2 × 10^-19 J
- D) 4 × 10^-19 J

Answer Key

Q1: C - Photoelectric current is directly proportional to light intensity.
Q2: B - Stopping potential is where no photoelectron reaches collector, current zero.
Q3: D - K_max = eV₀ = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J (corrected to option A).

Memory Aids

Stopping potential V₀: voltage to stop photoelectrons.
Photoelectric current ∝ light intensity.
Retarding potential reduces electron flow.

Setup and Working

Emitting electrode C (photosensitive) at negative potential; collecting electrode A at positive potential in evacuated tube.
Light of frequency ν > threshold frequency ν₀ causes emission of photoelectrons from C.
Photoelectrons travel to A, creating photoelectric current I proportional to number of emitted electrons.
If A is negative w.r.t C, retarding electric field reduces photoelectron flow, decreasing current.
Stopping potential V₀: retarding potential where photoelectric current becomes zero.

Solved Examples

Ex 1: Calculate stopping potential if max kinetic energy of photoelectrons is 2 × 10^-19 J.
- Use K_max = eV₀
- V₀ = K_max/e = (2 × 10^-19)/(1.6 × 10^-19) = 1.25 V

Practice Set

Q1: What happens to photoelectric current if light intensity increases?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q2: Stopping potential is the potential at which:
- A) Photoelectric current is maximum
- B) No photoelectron reaches collector
- C) Emission starts
- D) Intensity is maximum
Q3: If stopping potential is 2 V, max kinetic energy of photoelectrons is:
- A) 3.2 × 10^-19 J
- B) 1.6 × 10^-19 J
- C) 2 × 10^-19 J
- D) 4 × 10^-19 J

Answer Key

Q1: C - Photoelectric current is directly proportional to light intensity.
Q2: B - Stopping potential is where no photoelectron reaches collector, current zero.
Q3: D - K_max = eV₀ = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J (corrected to option A).

Memory Aids

Stopping potential V₀: voltage to stop photoelectrons.
Photoelectric current ∝ light intensity.
Retarding potential reduces electron flow.

Setup and Working

Emitting electrode C (photosensitive) at negative potential; collecting electrode A at positive potential in evacuated tube.
Light of frequency \( \nu \) > threshold frequency \( \nu_0 \) causes emission of photoelectrons from C.
Photoelectrons travel to A, creating photoelectric current \( I \) proportional to number of emitted electrons.
If A is negative w.r.t C, retarding electric field reduces photoelectron flow, decreasing current.
Stopping potential \( V_0 \): retarding potential where photoelectric current becomes zero.

Solved Examples

Ex 1: Calculate stopping potential if max kinetic energy of photoelectrons is \( 2 \times 10^{-19} \) J.
- Use \( K_{max} = eV_0 \)
- \( V_0 = \frac{K_{max}}{e} = \frac{2 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.25 \text{ V} \)

Practice Set

Q1: What happens to photoelectric current if light intensity increases?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q2: Stopping potential is the potential at which:
- A) Photoelectric current is maximum
- B) No photoelectron reaches collector
- C) Emission starts
- D) Intensity is maximum
Q3: If stopping potential is 2 V, max kinetic energy of photoelectrons is:
- A) \( 3.2 \times 10^{-19} \) J
- B) \( 1.6 \times 10^{-19} \) J
- C) \( 2 \times 10^{-19} \) J
- D) \( 4 \times 10^{-19} \) J

Answer Key

Q1: C - Photoelectric current is directly proportional to light intensity.
Q2: B - Stopping potential is where no photoelectron reaches collector, current zero.
Q3: D - \( K_{max} = eV_0 = 1.6 \times 10^{-19} \times 2 = 3.2 \times 10^{-19} \) J (corrected to option A).

Memory Aids

Stopping potential \( V_0 \): voltage to stop photoelectrons.
Photoelectric current \( \propto \) light intensity.
Retarding potential reduces electron flow.

Effect Intensity Photocurrent

Key Points

Photocurrent I is directly proportional to light intensity I ∝ Intensity.
Frequency and accelerating potential fixed; increasing photons increases intensity.
Graph shows linear increase of photocurrent with intensity.

Solved Examples

Ex 1: If intensity doubles, what happens to photocurrent?
- Photocurrent doubles (direct proportionality).

Practice Set

Q1: Photocurrent is proportional to:
- A) Frequency
- B) Intensity
- C) Wavelength
- D) Stopping potential
Q2: If intensity is tripled, photocurrent:
- A) Triples
- B) Halves
- C) Remains same
- D) Zero
Q3: Which is fixed when studying effect of intensity on photocurrent?
- A) Frequency and potential
- B) Intensity
- C) Stopping potential
- D) Photoelectron energy

Answer Key

Q1: B - Photocurrent ∝ Intensity.
Q2: A - Photocurrent triples with intensity.
Q3: A - Frequency and potential fixed to isolate intensity effect.

Memory Aids

Photocurrent ∝ Intensity (direct relation).
Frequency fixed during intensity variation.

Key Points

Photocurrent I is directly proportional to light intensity I ∝ Intensity.
Frequency and accelerating potential fixed; increasing photons increases intensity.
Graph shows linear increase of photocurrent with intensity.

Solved Examples

Ex 1: If intensity doubles, what happens to photocurrent?
- Photocurrent doubles (direct proportionality).

Practice Set

Q1: Photocurrent is proportional to:
- A) Frequency
- B) Intensity
- C) Wavelength
- D) Stopping potential
Q2: If intensity is tripled, photocurrent:
- A) Triples
- B) Halves
- C) Remains same
- D) Zero
Q3: Which is fixed when studying effect of intensity on photocurrent?
- A) Frequency and potential
- B) Intensity
- C) Stopping potential
- D) Photoelectron energy

Answer Key

Q1: B - Photocurrent ∝ Intensity.
Q2: A - Photocurrent triples with intensity.
Q3: A - Frequency and potential fixed to isolate intensity effect.

Memory Aids

Photocurrent ∝ Intensity (direct relation).
Frequency fixed during intensity variation.

Key Points

Photocurrent I is directly proportional to light intensity I ∝ Intensity.
Frequency and accelerating potential fixed; increasing photons increases intensity.
Graph shows linear increase of photocurrent with intensity.

Solved Examples

Ex 1: If intensity doubles, what happens to photocurrent?
- Photocurrent doubles (direct proportionality).

Practice Set

Q1: Photocurrent is proportional to:
- A) Frequency
- B) Intensity
- C) Wavelength
- D) Stopping potential
Q2: If intensity is tripled, photocurrent:
- A) Triples
- B) Halves
- C) Remains same
- D) Zero
Q3: Which is fixed when studying effect of intensity on photocurrent?
- A) Frequency and potential
- B) Intensity
- C) Stopping potential
- D) Photoelectron energy

Answer Key

Q1: B - Photocurrent ∝ Intensity.
Q2: A - Photocurrent triples with intensity.
Q3: A - Frequency and potential fixed to isolate intensity effect.

Memory Aids

Photocurrent ∝ Intensity (direct relation).
Frequency fixed during intensity variation.

Key Points

Photocurrent \( I \) is directly proportional to light intensity \( I \propto \text{Intensity} \).
Frequency and accelerating potential fixed; increasing photons increases intensity.
Graph shows linear increase of photocurrent with intensity.

Solved Examples

Ex 1: If intensity doubles, what happens to photocurrent?
- Photocurrent doubles (direct proportionality).

Practice Set

Q1: Photocurrent is proportional to:
- A) Frequency
- B) Intensity
- C) Wavelength
- D) Stopping potential
Q2: If intensity is tripled, photocurrent:
- A) Triples
- B) Halves
- C) Remains same
- D) Zero
Q3: Which is fixed when studying effect of intensity on photocurrent?
- A) Frequency and potential
- B) Intensity
- C) Stopping potential
- D) Photoelectron energy

Answer Key

Q1: B - Photocurrent ∝ Intensity.
Q2: A - Photocurrent triples with intensity.
Q3: A - Frequency and potential fixed to isolate intensity effect.

Memory Aids

Photocurrent ∝ Intensity (direct relation).
Frequency fixed during intensity variation.

Effect Potential Photoelectric Current

Key Points

Saturation current: max photoelectric current when all emitted electrons reach collector.
Saturation current depends on number of incident photons per second.
Photocurrent varies with collector plate potential and light intensity.
At stopping potential V₀, kinetic energy K_max = eV₀.
Stopping potential independent of light intensity for fixed frequency.

Solved Examples

Ex 1: Calculate max kinetic energy if stopping potential is 3 V.
- K_max = eV₀ = 1.6 × 10^-19 × 3 = 4.8 × 10^-19 J

Practice Set

Q1: Saturation current depends on:
- A) Frequency
- B) Number of photons
- C) Stopping potential
- D) Collector potential
Q2: Stopping potential is:
- A) Dependent on intensity
- B) Independent of intensity
- C) Zero always
- D) Maximum current
Q3: At stopping potential, kinetic energy is:
- A) Zero
- B) Equal to eV₀
- C) Maximum
- D) Negative

Answer Key

Q1: B - Saturation current depends on photon number.
Q2: B - Stopping potential independent of intensity.
Q3: B - K_max = eV₀ at stopping potential.

Memory Aids

Saturation current = max current.
Stopping potential independent of intensity.
K_max = eV₀ formula key.

Key Points

Saturation current: max photoelectric current when all emitted electrons reach collector.
Saturation current depends on number of incident photons per second.
Photocurrent varies with collector plate potential and light intensity.
At stopping potential V₀, kinetic energy K_max = eV₀.
Stopping potential independent of light intensity for fixed frequency.

Solved Examples

Ex 1: Calculate max kinetic energy if stopping potential is 3 V.
- K_max = eV₀ = 1.6 × 10^-19 × 3 = 4.8 × 10^-19 J

Practice Set

Q1: Saturation current depends on:
- A) Frequency
- B) Number of photons
- C) Stopping potential
- D) Collector potential
Q2: Stopping potential is:
- A) Dependent on intensity
- B) Independent of intensity
- C) Zero always
- D) Maximum current
Q3: At stopping potential, kinetic energy is:
- A) Zero
- B) Equal to eV₀
- C) Maximum
- D) Negative

Answer Key

Q1: B - Saturation current depends on photon number.
Q2: B - Stopping potential independent of intensity.
Q3: B - K_max = eV₀ at stopping potential.

Memory Aids

Saturation current = max current.
Stopping potential independent of intensity.
K_max = eV₀ formula key.

Key Points

Saturation current: max photoelectric current when all emitted electrons reach collector.
Saturation current depends on number of incident photons per second.
Photocurrent varies with collector plate potential and light intensity.
At stopping potential V₀, kinetic energy K_max = eV₀.
Stopping potential independent of light intensity for fixed frequency.

Solved Examples

Ex 1: Calculate max kinetic energy if stopping potential is 3 V.
- K_max = eV₀ = 1.6 × 10^-19 × 3 = 4.8 × 10^-19 J

Practice Set

Q1: Saturation current depends on:
- A) Frequency
- B) Number of photons
- C) Stopping potential
- D) Collector potential
Q2: Stopping potential is:
- A) Dependent on intensity
- B) Independent of intensity
- C) Zero always
- D) Maximum current
Q3: At stopping potential, kinetic energy is:
- A) Zero
- B) Equal to eV₀
- C) Maximum
- D) Negative

Answer Key

Q1: B - Saturation current depends on photon number.
Q2: B - Stopping potential independent of intensity.
Q3: B - K_max = eV₀ at stopping potential.

Memory Aids

Saturation current = max current.
Stopping potential independent of intensity.
K_max = eV₀ formula key.

Key Points

Saturation current: max photoelectric current when all emitted electrons reach collector.
Saturation current depends on number of incident photons per second.
Photocurrent varies with collector plate potential and light intensity.
At stopping potential \( V_0 \), kinetic energy \( K_{max} = eV_0 \).
Stopping potential independent of light intensity for fixed frequency.

Solved Examples

Ex 1: Calculate max kinetic energy if stopping potential is 3 V.
- \( K_{max} = eV_0 = 1.6 \times 10^{-19} \times 3 = 4.8 \times 10^{-19} \) J

Practice Set

Q1: Saturation current depends on:
- A) Frequency
- B) Number of photons
- C) Stopping potential
- D) Collector potential
Q2: Stopping potential is:
- A) Dependent on intensity
- B) Independent of intensity
- C) Zero always
- D) Maximum current
Q3: At stopping potential, kinetic energy is:
- A) Zero
- B) Equal to \( eV_0 \)
- C) Maximum
- D) Negative

Answer Key

Q1: B - Saturation current depends on photon number.
Q2: B - Stopping potential independent of intensity.
Q3: B - \( K_{max} = eV_0 \) at stopping potential.

Memory Aids

Saturation current = max current.
Stopping potential independent of intensity.
\( K_{max} = eV_0 \) formula key.

Effect Frequency Stopping Potential

Key Points

Stopping potential depends on frequency, not intensity.
Higher frequency ν₃ > ν₂ > ν₁ means higher stopping potential V₀₃ > V₀₂ > V₀₁.
Stopping potential varies linearly with frequency for given material.
Threshold frequency ν₀: minimum frequency for photoemission, where V₀ = 0.

Graph photocurrent vs potential for frequencies

Graph stopping potential vs frequency metals

Solved Examples

Ex 1: If threshold frequency is 5 × 10¹⁴ Hz, what happens below this frequency?
- No photoelectrons emitted, V₀ = 0.

Practice Set

Q1: Stopping potential depends on:
- A) Intensity
- B) Frequency
- C) Wavelength
- D) Number of photons
Q2: Threshold frequency is:
- A) Max frequency
- B) Min frequency for emission
- C) Frequency for max current
- D) Frequency for max intensity
Q3: For two metals A and B, if ν₀^A < ν₀^B, then:
- A) Metal A emits electrons at lower frequency
- B) Metal B emits electrons at lower frequency
- C) Both emit at same frequency
- D) No emission

Answer Key

Q1: B - Stopping potential depends on frequency.
Q2: B - Threshold frequency is minimum frequency for emission.
Q3: A - Metal with lower threshold frequency emits electrons first.

Memory Aids

Stopping potential V₀ linear with frequency.
Threshold frequency ν₀ is cut-off for emission.
Higher frequency → higher kinetic energy.

Key Points

Stopping potential depends on frequency, not intensity.
Higher frequency ν₃ > ν₂ > ν₁ means higher stopping potential V₀₃ > V₀₂ > V₀₁.
Stopping potential varies linearly with frequency for given material.
Threshold frequency ν₀: minimum frequency for photoemission, where V₀ = 0.

Solved Examples

Ex 1: If threshold frequency is 5 × 10¹⁴ Hz, what happens below this frequency?
- No photoelectrons emitted, V₀ = 0.

Practice Set

Q1: Stopping potential depends on:
- A) Intensity
- B) Frequency
- C) Wavelength
- D) Number of photons
Q2: Threshold frequency is:
- A) Max frequency
- B) Min frequency for emission
- C) Frequency for max current
- D) Frequency for max intensity
Q3: For two metals A and B, if ν₀^A < ν₀^B, then:
- A) Metal A emits electrons at lower frequency
- B) Metal B emits electrons at lower frequency
- C) Both emit at same frequency
- D) No emission

Answer Key

Q1: B - Stopping potential depends on frequency.
Q2: B - Threshold frequency is minimum frequency for emission.
Q3: A - Metal with lower threshold frequency emits electrons first.

Memory Aids

Stopping potential V₀ linear with frequency.
Threshold frequency ν₀ is cut-off for emission.
Higher frequency → higher kinetic energy.

Key Points

Stopping potential depends on frequency, not intensity.
Higher frequency ν₃ > ν₂ > ν₁ means higher stopping potential V₀₃ > V₀₂ > V₀₁.
Stopping potential varies linearly with frequency for given material.
Threshold frequency ν₀: minimum frequency for photoemission, where V₀ = 0.

Solved Examples

Ex 1: If threshold frequency is 5 × 10¹⁴ Hz, what happens below this frequency?
- No photoelectrons emitted, V₀ = 0.

Practice Set

Q1: Stopping potential depends on:
- A) Intensity
- B) Frequency
- C) Wavelength
- D) Number of photons
Q2: Threshold frequency is:
- A) Max frequency
- B) Min frequency for emission
- C) Frequency for max current
- D) Frequency for max intensity
Q3: For two metals A and B, if ν₀^A < ν₀^B, then:
- A) Metal A emits electrons at lower frequency
- B) Metal B emits electrons at lower frequency
- C) Both emit at same frequency
- D) No emission

Answer Key

Q1: B - Stopping potential depends on frequency.
Q2: B - Threshold frequency is minimum frequency for emission.
Q3: A - Metal with lower threshold frequency emits electrons first.

Memory Aids

Stopping potential V₀ linear with frequency.
Threshold frequency ν₀ is cut-off for emission.
Higher frequency → higher kinetic energy.

Key Points

Stopping potential depends on frequency, not intensity.
Higher frequency \( \nu_3 > \nu_2 > \nu_1 \) means higher stopping potential \( V_{03} > V_{02} > V_{01} \).
Stopping potential varies linearly with frequency for given material.
Threshold frequency \( \nu_0 \): minimum frequency for photoemission, where \( V_0 = 0 \).

Solved Examples

Ex 1: If threshold frequency is \( 5 \times 10^{14} \) Hz, what happens below this frequency?
- No photoelectrons emitted, \( V_0 = 0 \).

Practice Set

Q1: Stopping potential depends on:
- A) Intensity
- B) Frequency
- C) Wavelength
- D) Number of photons
Q2: Threshold frequency is:
- A) Max frequency
- B) Min frequency for emission
- C) Frequency for max current
- D) Frequency for max intensity
Q3: For two metals A and B, if \( \nu_0^A < \nu_0^B \), then:
- A) Metal A emits electrons at lower frequency
- B) Metal B emits electrons at lower frequency
- C) Both emit at same frequency
- D) No emission

Answer Key

Q1: B - Stopping potential depends on frequency.
Q2: B - Threshold frequency is minimum frequency for emission.
Q3: A - Metal with lower threshold frequency emits electrons first.

Memory Aids

Stopping potential \( V_0 \) linear with frequency.
Threshold frequency \( \nu_0 \) is cut-off for emission.
Higher frequency → higher kinetic energy.

Photoelectric Effect and Wave Theory of Light

Wave Theory of Light

Light is an electromagnetic wave with electric and magnetic fields oscillating perpendicular to each other.
Energy is continuously distributed over the wave's spatial extent.
Intensity of light relates to the amplitude of electric and magnetic fields.

Wave Theory Explanation of Photoelectric Effect

Electrons on metal surface absorb energy continuously from incident light.
Higher intensity means larger amplitude fields, so electrons absorb more energy.
Expected: Maximum kinetic energy of photoelectrons increases with intensity.
Expected: Any frequency of light, if intense enough, should eject electrons after some time.
Expected: No threshold frequency should exist.
These predictions contradict experimental observations.

Limitations of Wave Theory

Cannot explain the existence of threshold frequency.
Cannot explain why kinetic energy depends on frequency, not intensity.
Fails to describe instantaneous emission of photoelectrons.

Solved Examples

Ex 1: If intensity of light doubles, what happens to maximum kinetic energy of photoelectrons according to wave theory?
Answer: It should double (incorrect as per experiments).
Ex 2: For light of frequency below threshold, can electrons be emitted by increasing intensity?
Answer: Yes, wave theory predicts emission after some time (contradicted by experiments).

Practice Set

Q1: According to wave theory, increasing light intensity will:
A) Increase number of photoelectrons only
B) Increase kinetic energy of photoelectrons only
C) Increase both number and kinetic energy
D) Have no effect
Q2: Wave theory predicts threshold frequency to be:
A) Zero
B) Equal to frequency of light
C) Dependent on intensity
D) Infinite
Q3: Which observation contradicts wave theory?
A) Photoelectrons emitted instantaneously
B) Kinetic energy depends on frequency
C) No electrons emitted below threshold frequency
D) All of the above

Answer Key

Q1: A - Intensity increases number of electrons but wave theory incorrectly predicts kinetic energy increase.
Q2: A - Wave theory predicts no threshold frequency (effectively zero), contradicting experiments.
Q3: D - All listed observations contradict wave theory.

Memory Aids

Wave theory: Energy intensity, not frequency.
Photoelectric effect: Threshold frequency exists.
Instant emission particle nature of light.

Wave Theory of Light

Light is an electromagnetic wave with electric and magnetic fields oscillating perpendicular to each other.
Energy is continuously distributed over the wave's spatial extent.
Intensity of light relates to the amplitude of electric and magnetic fields.

Wave Theory Explanation of Photoelectric Effect

Electrons on metal surface absorb energy continuously from incident light.
Higher intensity means larger amplitude fields, so electrons absorb more energy.
Expected: Maximum kinetic energy of photoelectrons increases with intensity.
Expected: Any frequency of light, if intense enough, should eject electrons after some time.
Expected: No threshold frequency should exist.
These predictions contradict experimental observations.

Limitations of Wave Theory

Cannot explain the existence of threshold frequency.
Cannot explain why kinetic energy depends on frequency, not intensity.
Fails to describe instantaneous emission of photoelectrons.

Solved Examples

Ex 1: If intensity of light doubles, what happens to maximum kinetic energy of photoelectrons according to wave theory?
Answer: It should double (incorrect as per experiments).
Ex 2: For light of frequency below threshold, can electrons be emitted by increasing intensity?
Answer: Yes, wave theory predicts emission after some time (contradicted by experiments).

Practice Set

Q1: According to wave theory, increasing light intensity will:
A) Increase number of photoelectrons only
B) Increase kinetic energy of photoelectrons only
C) Increase both number and kinetic energy
D) Have no effect
Q2: Wave theory predicts threshold frequency to be:
A) Zero
B) Equal to frequency of light
C) Dependent on intensity
D) Infinite
Q3: Which observation contradicts wave theory?
A) Photoelectrons emitted instantaneously
B) Kinetic energy depends on frequency
C) No electrons emitted below threshold frequency
D) All of the above

Answer Key

Q1: A - Intensity increases number of electrons but wave theory incorrectly predicts kinetic energy increase.
Q2: A - Wave theory predicts no threshold frequency (effectively zero), contradicting experiments.
Q3: D - All listed observations contradict wave theory.

Memory Aids

Wave theory: Energy intensity, not frequency.
Photoelectric effect: Threshold frequency exists.
Instant emission particle nature of light.

Wave Theory of Light

Light is an electromagnetic wave with electric and magnetic fields oscillating perpendicular to each other.
Energy is continuously distributed over the wave's spatial extent.
Intensity of light relates to the amplitude of electric and magnetic fields.

Wave Theory Explanation of Photoelectric Effect

Electrons on metal surface absorb energy continuously from incident light.
Higher intensity means larger amplitude fields, so electrons absorb more energy.
Expected: Maximum kinetic energy of photoelectrons increases with intensity.
Expected: Any frequency of light, if intense enough, should eject electrons after some time.
Expected: No threshold frequency should exist.
These predictions contradict experimental observations.

Limitations of Wave Theory

Cannot explain the existence of threshold frequency.
Cannot explain why kinetic energy depends on frequency, not intensity.
Fails to describe instantaneous emission of photoelectrons.

Solved Examples

Ex 1: If intensity of light doubles, what happens to maximum kinetic energy of photoelectrons according to wave theory?
Answer: It should double (incorrect as per experiments).
Ex 2: For light of frequency below threshold, can electrons be emitted by increasing intensity?
Answer: Yes, wave theory predicts emission after some time (contradicted by experiments).

Practice Set

Q1: According to wave theory, increasing light intensity will:
A) Increase number of photoelectrons only
B) Increase kinetic energy of photoelectrons only
C) Increase both number and kinetic energy
D) Have no effect
Q2: Wave theory predicts threshold frequency to be:
A) Zero
B) Equal to frequency of light
C) Dependent on intensity
D) Infinite
Q3: Which observation contradicts wave theory?
A) Photoelectrons emitted instantaneously
B) Kinetic energy depends on frequency
C) No electrons emitted below threshold frequency
D) All of the above

Answer Key

Q1: A - Intensity increases number of electrons but wave theory incorrectly predicts kinetic energy increase.
Q2: A - Wave theory predicts no threshold frequency (effectively zero), contradicting experiments.
Q3: D - All listed observations contradict wave theory.

Memory Aids

Wave theory: Energy intensity, not frequency.
Photoelectric effect: Threshold frequency exists.
Instant emission particle nature of light.

Wave Theory of Light

Light is an electromagnetic wave with electric and magnetic fields oscillating perpendicular to each other.
Energy is continuously distributed over the wave's spatial extent.
Intensity of light relates to the amplitude of electric and magnetic fields.

Wave Theory Explanation of Photoelectric Effect

Electrons on metal surface absorb energy continuously from incident light.
Higher intensity means larger amplitude fields, so electrons absorb more energy.
Expected: Maximum kinetic energy of photoelectrons increases with intensity.
Expected: Any frequency of light, if intense enough, should eject electrons after some time.
Expected: No threshold frequency should exist.
These predictions contradict experimental observations.

Limitations of Wave Theory

Cannot explain the existence of threshold frequency.
Cannot explain why kinetic energy depends on frequency, not intensity.
Fails to describe instantaneous emission of photoelectrons.

Solved Examples

Ex 1: If intensity of light doubles, what happens to maximum kinetic energy of photoelectrons according to wave theory?
Answer: It should double (incorrect as per experiments).
Ex 2: For light of frequency below threshold, can electrons be emitted by increasing intensity?
Answer: Yes, wave theory predicts emission after some time (contradicted by experiments).

Practice Set

Q1: According to wave theory, increasing light intensity will:
A) Increase number of photoelectrons only
B) Increase kinetic energy of photoelectrons only
C) Increase both number and kinetic energy
D) Have no effect
Q2: Wave theory predicts threshold frequency to be:
A) Zero
B) Equal to frequency of light
C) Dependent on intensity
D) Infinite
Q3: Which observation contradicts wave theory?
A) Photoelectrons emitted instantaneously
B) Kinetic energy depends on frequency
C) No electrons emitted below threshold frequency
D) All of the above

Answer Key

Q1: A - Intensity increases number of electrons but wave theory incorrectly predicts kinetic energy increase.
Q2: A - Wave theory predicts no threshold frequency (effectively zero), contradicting experiments.
Q3: D - All listed observations contradict wave theory.

Memory Aids

Wave theory: Energy intensity, not frequency.
Photoelectric effect: Threshold frequency exists.
Instant emission particle nature of light.

Einstein's Photoelectric Equation

Explanation

When light of frequency ν hits a metal, photons transfer energy hν to electrons.
Energy used to overcome work function W₀ and rest becomes kinetic energy of emitted electron.
Photoelectric equation: hν = 1/2mv_max² + W₀
At threshold frequency ν₀, W₀ = hν₀ = hc/λ₀
Maximum kinetic energy: 1/2mv_max² = h(ν - ν₀)
Light intensity affects number of electrons, not their energy.
Stopping potential depends on frequency, not intensity.

Diagram 1 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Explanation

When light of frequency ν hits a metal, photons transfer energy hν to electrons.
Energy used to overcome work function W₀ and rest becomes kinetic energy of emitted electron.
Photoelectric equation: hν = 1/2mv_max² + W₀
At threshold frequency ν₀, W₀ = hν₀ = hc/λ₀
Maximum kinetic energy: 1/2mv_max² = h(ν - ν₀)
Light intensity affects number of electrons, not their energy.
Stopping potential depends on frequency, not intensity.

Diagram 2 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Explanation

When light of frequency ν hits a metal, photons transfer energy hν to electrons.
Energy used to overcome work function W₀ and rest becomes kinetic energy of emitted electron.
Photoelectric equation: hν = 1/2mv_max² + W₀
At threshold frequency ν₀, W₀ = hν₀ = hc/λ₀
Maximum kinetic energy: 1/2mv_max² = h(ν - ν₀)
Light intensity affects number of electrons, not their energy.
Stopping potential depends on frequency, not intensity.

Diagram 3 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Calculate maximum kinetic energy if ν = 1.5 × 10¹⁵ Hz, ν₀ = 1 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js.
- K_max = h(ν - ν₀) = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J

Practice Set

Q1: What is the threshold frequency ν₀ related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency 2 × 10¹⁵ Hz and threshold frequency 1.5 × 10¹⁵ Hz, h = 6.63 × 10^-34 Js, calculate K_max.
- A) 3.315 × 10^-19 J
- B) 6.63 × 10^-19 J
- C) 9.945 × 10^-19 J
- D) 1.33 × 10^-19 J

Answer Key

Q1: B - Threshold frequency corresponds to work function W₀ = hν₀.
Q2: C - Kinetic energy increases as K_max = h(ν - ν₀).
Q3: A - K_max = 6.63 × 10^-34 × (0.5 × 10¹⁵) = 3.315 × 10^-19 J.

Memory Aids

Photoelectric eq: hν = K.E. + W₀
Threshold freq: ν₀ = W₀/h
Kinetic energy depends on ν - ν₀

Explanation

When light of frequency \( \nu \) hits a metal, photons transfer energy \( h\nu \) to electrons.
Energy used to overcome work function \( W_0 \) and rest becomes kinetic energy of emitted electron.
Photoelectric equation: \( h\nu = \frac{1}{2}mv_{max}^2 + W_0 \)
At threshold frequency \( \nu_0 \), \( W_0 = h\nu_0 = \frac{hc}{\lambda_0} \)
Maximum kinetic energy: \( \frac{1}{2}mv_{max}^2 = h(\nu - \nu_0) \)
Light intensity affects number of electrons, not their energy.
Stopping potential depends on frequency, not intensity.

Diagram 4 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Calculate maximum kinetic energy if \( \nu = 1.5 \times 10^{15} \) Hz, \( \nu_0 = 1 \times 10^{15} \) Hz, \( h = 6.63 \times 10^{-34} \) Js.
- \( K_{max} = h(\nu - \nu_0) = 6.63 \times 10^{-34} \times (0.5 \times 10^{15}) = 3.315 \times 10^{-19} \) J

Practice Set

Q1: What is the threshold frequency \( \nu_0 \) related to?
- A) Maximum kinetic energy
- B) Work function
- C) Photon energy
- D) Stopping potential
Q2: If frequency of light increases, what happens to the kinetic energy of emitted electrons?
- A) Decreases
- B) Remains same
- C) Increases
- D) Becomes zero
Q3: For light of frequency \( 2 \times 10^{15} \) Hz and threshold frequency \( 1.5 \times 10^{15} \) Hz, \( h = 6.63 \times 10^{-34} \) Js, calculate \( K_{max} \).
- A) \( 3.315 \times 10^{-19} \) J
- B) \( 6.63 \times 10^{-19} \) J
- C) \( 9.945 \times 10^{-19} \) J
- D) \( 1.33 \times 10^{-19} \) J

Answer Key

Q1: B - Threshold frequency corresponds to work function \( W_0 = h\nu_0 \).
Q2: C - Kinetic energy increases as \( K_{max} = h(\nu - \nu_0) \).
Q3: A - \( K_{max} = 6.63 \times 10^{-34} \times (0.5 \times 10^{15}) = 3.315 \times 10^{-19} \) J.

Memory Aids

Photoelectric eq: \( h\nu = K.E. + W_0 \)
Threshold freq: \( \nu_0 = W_0/h \)
Kinetic energy depends on \( \nu - \nu_0 \)

Determination of Planck's Constant and Work Function

Explanation

From photoelectric equation: K_max = hν - W₀
Stopping potential V₀ relates as eV₀ = hν - W₀
Rearranged: V₀ = (h/e) ν - W₀/e = (h/e) ν - hν₀/e
Plot of V₀ vs ν is a straight line with slope h/e
Slope = tan θ = V₀/(ν - ν₀) = h/e
Intercept on potential axis: -hν₀/e
Work function: W₀ = hν₀ = e × OA
Planck's constant h can be found from slope h/e

Diagram 5 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Explanation

From photoelectric equation: K_max = hν - W₀
Stopping potential V₀ relates as eV₀ = hν - W₀
Rearranged: V₀ = (h/e) ν - W₀/e = (h/e) ν - hν₀/e
Plot of V₀ vs ν is a straight line with slope h/e
Slope = tan θ = V₀/(ν - ν₀) = h/e
Intercept on potential axis: -hν₀/e
Work function: W₀ = hν₀ = e × OA
Planck's constant h can be found from slope h/e

Diagram 6 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Explanation

From photoelectric equation: K_max = hν - W₀
Stopping potential V₀ relates as eV₀ = hν - W₀
Rearranged: V₀ = (h/e) ν - W₀/e = (h/e) ν - hν₀/e
Plot of V₀ vs ν is a straight line with slope h/e
Slope = tan θ = V₀/(ν - ν₀) = h/e
Intercept on potential axis: -hν₀/e
Work function: W₀ = hν₀ = e × OA
Planck's constant h can be found from slope h/e

Diagram 7 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Given slope = 4.14 × 10^-15 V·s, find Planck's constant h (electron charge e = 1.6 × 10^-19 C).
- h = slope × e = 4.14 × 10^-15 × 1.6 × 10^-19 = 6.62 × 10^-34 Js
Ex 2: If intercept on potential axis is -2 V, find work function W₀.
- W₀ = e × |intercept| = 1.6 × 10^-19 × 2 = 3.2 × 10^-19 J

Practice Set

Q1: What does the slope of V₀ vs ν graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential V₀ = 1.5 V at frequency 5 × 10¹⁴ Hz, and W₀ = 2 × 10^-19 J, find h (e = 1.6 × 10^-19 C).
- A) 4.8 × 10^-34 Js
- B) 6.63 × 10^-34 Js
- C) 3.3 × 10^-34 Js
- D) 1.2 × 10^-34 Js
Q3: The intercept on potential axis is -3 V, electron charge e = 1.6 × 10^-19 C. Calculate work function W₀.
- A) 4.8 × 10^-19 J
- B) 3.2 × 10^-19 J
- C) 1.6 × 10^-19 J
- D) 6.4 × 10^-19 J

Answer Key

Q1: B - Slope = h/e relates Planck's constant and electron charge.
Q2: A - Use V₀ = (h/e) ν - W₀/e to find h.
Q3: A - W₀ = e × 3 = 4.8 × 10^-19 J.

Memory Aids

Slope of V₀ vs ν = h/e
Intercept = -hν₀/e
Work function W₀ = e × |intercept|

Explanation

From photoelectric equation: \( K_{max} = h\nu - W_0 \)
Stopping potential \( V_0 \) relates as \( eV_0 = h\nu - W_0 \)
Rearranged: \( V_0 = \frac{h}{e} \nu - \frac{W_0}{e} = \frac{h}{e} \nu - \frac{h\nu_0}{e} \)
Plot of \( V_0 \) vs \( \nu \) is a straight line with slope \( \frac{h}{e} \)
Slope \( = \tan \theta = \frac{V_0}{\nu - \nu_0} = \frac{h}{e} \)
Intercept on potential axis: \( -\frac{h\nu_0}{e} \)
Work function: \( W_0 = h\nu_0 = e \times OA \)
Planck's constant \( h \) can be found from slope \( \frac{h}{e} \)

Diagram 8 – NEET Physics Dual Nature of Radiation and Matter

Solved Examples

Ex 1: Given slope \( = 4.14 \times 10^{-15} \) V·s, find Planck's constant \( h \) (electron charge \( e = 1.6 \times 10^{-19} \) C).
- \( h = slope \times e = 4.14 \times 10^{-15} \times 1.6 \times 10^{-19} = 6.62 \times 10^{-34} \) Js
Ex 2: If intercept on potential axis is \( -2 \) V, find work function \( W_0 \).
- \( W_0 = e \times |intercept| = 1.6 \times 10^{-19} \times 2 = 3.2 \times 10^{-19} \) J

Practice Set

Q1: What does the slope of \( V_0 \) vs \( \nu \) graph represent?
- A) Work function
- B) Planck's constant divided by electron charge
- C) Threshold frequency
- D) Maximum kinetic energy
Q2: If stopping potential \( V_0 = 1.5 \) V at frequency \( 5 \times 10^{14} \) Hz, and \( W_0 = 2 \times 10^{-19} \) J, find \( h \) (\( e = 1.6 \times 10^{-19} \) C).
- A) \( 4.8 \times 10^{-34} \) Js
- B) \( 6.63 \times 10^{-34} \) Js
- C) \( 3.3 \times 10^{-34} \) Js
- D) \( 1.2 \times 10^{-34} \) Js
Q3: The intercept on potential axis is \( -3 \) V, electron charge \( e = 1.6 \times 10^{-19} \) C. Calculate work function \( W_0 \).
- A) \( 4.8 \times 10^{-19} \) J
- B) \( 3.2 \times 10^{-19} \) J
- C) \( 1.6 \times 10^{-19} \) J
- D) \( 6.4 \times 10^{-19} \) J

Answer Key

Q1: B - Slope = \( \frac{h}{e} \) relates Planck's constant and electron charge.
Q2: A - Use \( V_0 = \frac{h}{e} \nu - \frac{W_0}{e} \) to find \( h \).
Q3: A - \( W_0 = e \times 3 = 4.8 \times 10^{-19} \) J.

Memory Aids

Slope of \( V_0 \) vs \( \nu \) = \( h/e \)
Intercept = \( -h\nu_0/e \)
Work function \( W_0 = e \times |intercept| \)

Particle Nature of Light : The Photon

Photon Concept and Properties

Einstein proposed electromagnetic radiation is quantized into photons.
Radiation behaves as particles (photons) in matter interaction.
Photon energy: E = hν = hc/λ, where h = Planck's constant, ν = frequency, λ = wavelength, c = speed of light.
Photon momentum: p = hν/c = h/λ.
Photon energy is independent of radiation intensity.
All photons travel at speed c in vacuum.
Photon frequency determines energy (color), unchanged in different media.
Photons are electrically neutral; not deflected by electric/magnetic fields.
Photons are massless; rest mass = 0.
Energy and momentum conserved in photon-particle collisions; photon number may change.

Solved Examples

Ex 1: Calculate photon energy for wavelength 500 nm.
- Use E = hc/λ, h = 6.63×10^-34 Js, c = 3×10⁸ m/s, λ = 500×10^-9 m.
- Calculate E = (6.63×10^-34 × 3×10⁸)/(500×10^-9) = 3.98×10^-19 J.
Ex 2: Find photon momentum for wavelength 600 nm.
- Use p = h/λ.
- Calculate p = 6.63×10^-34/(600×10^-9) = 1.11×10^-27 kg·m/s.

Practice Set

Q1: What is the energy of a photon with frequency 5×10¹⁴ Hz?
- A) 3.32×10^-19 J
- B) 6.63×10^-34 J
- C) 1.5×10^-15 J
- D) 9.11×10^-31 J
Q2: Which property of photons shows they are electrically neutral?
- A) Deflection by magnetic fields
- B) Deflection by electric fields
- C) No deflection by electric or magnetic fields
- D) Having rest mass zero
Q3: Photon momentum for wavelength 400 nm is?
- A) 1.66×10^-27 kg·m/s
- B) 4.97×10^-28 kg·m/s
- C) 6.63×10^-34 kg·m/s
- D) 3.32×10^-19 kg·m/s

Answer Key

Q1: A) 3.32×10^-19 J - Calculated using E = hν with given frequency.
Q2: C) No deflection by electric or magnetic fields - shows photons are neutral.
Q3: B) 4.97×10^-28 kg·m/s - Using p = h/λ with λ = 400 nm.

Memory Aids

Energy formula: E = hν = hc/λ
Momentum formula: p = h/λ
Photons: massless, neutral, speed = c
Energy independent of intensity
Frequency fixes photon energy (color)

Photon Concept and Properties

Einstein proposed electromagnetic radiation is quantized into photons.
Radiation behaves as particles (photons) in matter interaction.
Photon energy: E = hν = hc/λ, where h = Planck's constant, ν = frequency, λ = wavelength, c = speed of light.
Photon momentum: p = hν/c = h/λ.
Photon energy is independent of radiation intensity.
All photons travel at speed c in vacuum.
Photon frequency determines energy (color), unchanged in different media.
Photons are electrically neutral; not deflected by electric/magnetic fields.
Photons are massless; rest mass = 0.
Energy and momentum conserved in photon-particle collisions; photon number may change.

Solved Examples

Ex 1: Calculate photon energy for wavelength 500 nm.
- Use E = hc/λ, h = 6.63×10^-34 Js, c = 3×10⁸ m/s, λ = 500×10^-9 m.
- Calculate E = (6.63×10^-34 × 3×10⁸)/(500×10^-9) = 3.98×10^-19 J.
Ex 2: Find photon momentum for wavelength 600 nm.
- Use p = h/λ.
- Calculate p = 6.63×10^-34/(600×10^-9) = 1.11×10^-27 kg·m/s.

Practice Set

Q1: What is the energy of a photon with frequency 5×10¹⁴ Hz?
- A) 3.32×10^-19 J
- B) 6.63×10^-34 J
- C) 1.5×10^-15 J
- D) 9.11×10^-31 J
Q2: Which property of photons shows they are electrically neutral?
- A) Deflection by magnetic fields
- B) Deflection by electric fields
- C) No deflection by electric or magnetic fields
- D) Having rest mass zero
Q3: Photon momentum for wavelength 400 nm is?
- A) 1.66×10^-27 kg·m/s
- B) 4.97×10^-28 kg·m/s
- C) 6.63×10^-34 kg·m/s
- D) 3.32×10^-19 kg·m/s

Answer Key

Q1: A) 3.32×10^-19 J - Calculated using E = hν with given frequency.
Q2: C) No deflection by electric or magnetic fields - shows photons are neutral.
Q3: B) 4.97×10^-28 kg·m/s - Using p = h/λ with λ = 400 nm.

Memory Aids

Energy formula: E = hν = hc/λ
Momentum formula: p = h/λ
Photons: massless, neutral, speed = c
Energy independent of intensity
Frequency fixes photon energy (color)

Photon Concept and Properties

Einstein proposed electromagnetic radiation is quantized into photons.
Radiation behaves as particles (photons) in matter interaction.
Photon energy: E = hν = hc/λ, where h = Planck's constant, ν = frequency, λ = wavelength, c = speed of light.
Photon momentum: p = hν/c = h/λ.
Photon energy is independent of radiation intensity.
All photons travel at speed c in vacuum.
Photon frequency determines energy (color), unchanged in different media.
Photons are electrically neutral; not deflected by electric/magnetic fields.
Photons are massless; rest mass = 0.
Energy and momentum conserved in photon-particle collisions; photon number may change.

Solved Examples

Ex 1: Calculate photon energy for wavelength 500 nm.
- Use E = hc/λ, h = 6.63×10^-34 Js, c = 3×10⁸ m/s, λ = 500×10^-9 m.
- Calculate E = (6.63×10^-34 × 3×10⁸)/(500×10^-9) = 3.98×10^-19 J.
Ex 2: Find photon momentum for wavelength 600 nm.
- Use p = h/λ.
- Calculate p = 6.63×10^-34/(600×10^-9) = 1.11×10^-27 kg·m/s.

Practice Set

Q1: What is the energy of a photon with frequency 5×10¹⁴ Hz?
- A) 3.32×10^-19 J
- B) 6.63×10^-34 J
- C) 1.5×10^-15 J
- D) 9.11×10^-31 J
Q2: Which property of photons shows they are electrically neutral?
- A) Deflection by magnetic fields
- B) Deflection by electric fields
- C) No deflection by electric or magnetic fields
- D) Having rest mass zero
Q3: Photon momentum for wavelength 400 nm is?
- A) 1.66×10^-27 kg·m/s
- B) 4.97×10^-28 kg·m/s
- C) 6.63×10^-34 kg·m/s
- D) 3.32×10^-19 kg·m/s

Answer Key

Q1: A) 3.32×10^-19 J - Calculated using E = hν with given frequency.
Q2: C) No deflection by electric or magnetic fields - shows photons are neutral.
Q3: B) 4.97×10^-28 kg·m/s - Using p = h/λ with λ = 400 nm.

Memory Aids

Energy formula: E = hν = hc/λ
Momentum formula: p = h/λ
Photons: massless, neutral, speed = c
Energy independent of intensity
Frequency fixes photon energy (color)

Photon Concept and Properties

Einstein proposed electromagnetic radiation is quantized into photons.
Radiation behaves as particles (photons) in matter interaction.
Photon energy: E = h\nu = \frac{hc}{\lambda}, where h = Planck's constant, \nu = frequency, \lambda = wavelength, c = speed of light.
Photon momentum: p = \frac{h\nu}{c} = \frac{h}{\lambda}.
Photon energy is independent of radiation intensity.
All photons travel at speed c in vacuum.
Photon frequency determines energy (color), unchanged in different media.
Photons are electrically neutral; not deflected by electric/magnetic fields.
Photons are massless; rest mass = 0.
Energy and momentum conserved in photon-particle collisions; photon number may change.

Solved Examples

Ex 1: Calculate photon energy for wavelength 500 nm.
- Use E = \frac{hc}{\lambda}, h = 6.63×10^-34 Js, c = 3×10⁸ m/s, \lambda = 500×10^-9 m.
- Calculate E = \frac{6.63\times10^{-34} \times 3\times10^{8}}{500\times10^{-9}} = 3.98\times10^{-19} J.
Ex 2: Find photon momentum for wavelength 600 nm.
- Use p = \frac{h}{\lambda}.
- Calculate p = \frac{6.63\times10^{-34}}{600\times10^{-9}} = 1.11\times10^{-27} kg·m/s.

Practice Set

Q1: What is the energy of a photon with frequency 5×10¹⁴ Hz?
- A) 3.32×10^-19 J
- B) 6.63×10^-34 J
- C) 1.5×10^-15 J
- D) 9.11×10^-31 J
Q2: Which property of photons shows they are electrically neutral?
- A) Deflection by magnetic fields
- B) Deflection by electric fields
- C) No deflection by electric or magnetic fields
- D) Having rest mass zero
Q3: Photon momentum for wavelength 400 nm is?
- A) 1.66×10^-27 kg·m/s
- B) 4.97×10^-28 kg·m/s
- C) 6.63×10^-34 kg·m/s
- D) 3.32×10^-19 kg·m/s

Answer Key

Q1: A) 3.32×10^-19 J - Calculated using E = h\nu with given frequency.
Q2: C) No deflection by electric or magnetic fields - shows photons are neutral.
Q3: B) 4.97×10^-28 kg·m/s - Using p = h/\lambda with \lambda = 400 nm.

Memory Aids

Energy formula: E = h\nu = \frac{hc}{\lambda}
Momentum formula: p = \frac{h}{\lambda}
Photons: massless, neutral, speed = c
Energy independent of intensity
Frequency fixes photon energy (color)

Wave Nature of Matter

Diagram 9 – NEET Physics Dual Nature of Radiation and Matter

Concepts and Formulas

Every moving particle exhibits wave-like properties (de Broglie hypothesis).
De Broglie wavelength λ depends on particle momentum p = mv.
Formula: λ = h/p = h/mv, where h = 6.626 × 10^-34 Js.
For an electron accelerated by voltage V, λ = h/√(2meV) = 1.227/√V nm.
Wave-particle duality applies to matter similar to radiation.
De Broglie waves are also called matter waves.
Louis de Broglie received the Nobel Prize in 1929 for this discovery.

Solved Examples

Ex 1: Calculate de Broglie wavelength of an electron moving at speed 2 × 10⁶ m/s.
- Given: m_e = 9.11 × 10^-31 kg, v = 2 × 10⁶ m/s, h = 6.626 × 10^-34 Js
- λ = h/mv = 6.626 × 10^-34/(9.11 × 10^-31 × 2 × 10⁶) = 3.64 × 10^-10 m = 0.364 nm
Ex 2: Find λ for electron accelerated by V = 100 V.
- λ = 1.227/√V = 1.227/√100 = 0.1227 nm

Practice Set

Q1: What is the de Broglie wavelength of a particle with mass 2 × 10^-27 kg moving at 3 × 10³ m/s?
- A) 1.1 × 10^-34 m
- B) 1.1 × 10^-31 m
- C) 1.1 × 10^-37 m
- D) 1.1 × 10^-28 m
Q2: An electron accelerated through 400 V has de Broglie wavelength approximately:
- A) 0.061 nm
- B) 0.122 nm
- C) 0.031 nm
- D) 0.244 nm
Q3: If the velocity of a particle doubles, its de Broglie wavelength:
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: A) 1.1 × 10^-34 m - Using λ = h/mv, substituting values gives this wavelength.
Q2: A) 0.061 nm - λ = 1.227/√400 = 0.061 nm.
Q3: B) Halves - λ ∝ 1/v, doubling velocity halves wavelength.

Memory Aids

De Broglie wavelength λ = h/mv (momentum inverse relation).
Electron wavelength λ = 1.227/√V nm for accelerating voltage V.
Wave-particle duality: matter behaves like waves when moving.

Concepts and Formulas

Every moving particle exhibits wave-like properties (de Broglie hypothesis).
De Broglie wavelength λ depends on particle momentum p = mv.
Formula: λ = h/p = h/mv, where h = 6.626 × 10^-34 Js.
For an electron accelerated by voltage V, λ = h/√(2meV) = 1.227/√V nm.
Wave-particle duality applies to matter similar to radiation.
De Broglie waves are also called matter waves.
Louis de Broglie received the Nobel Prize in 1929 for this discovery.

Solved Examples

Ex 1: Calculate de Broglie wavelength of an electron moving at speed 2 × 10⁶ m/s.
- Given: m_e = 9.11 × 10^-31 kg, v = 2 × 10⁶ m/s, h = 6.626 × 10^-34 Js
- λ = h/mv = 6.626 × 10^-34/(9.11 × 10^-31 × 2 × 10⁶) = 3.64 × 10^-10 m = 0.364 nm
Ex 2: Find λ for electron accelerated by V = 100 V.
- λ = 1.227/√V = 1.227/√100 = 0.1227 nm

Practice Set

Q1: What is the de Broglie wavelength of a particle with mass 2 × 10^-27 kg moving at 3 × 10³ m/s?
- A) 1.1 × 10^-34 m
- B) 1.1 × 10^-31 m
- C) 1.1 × 10^-37 m
- D) 1.1 × 10^-28 m
Q2: An electron accelerated through 400 V has de Broglie wavelength approximately:
- A) 0.061 nm
- B) 0.122 nm
- C) 0.031 nm
- D) 0.244 nm
Q3: If the velocity of a particle doubles, its de Broglie wavelength:
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: A) 1.1 × 10^-34 m - Using λ = h/mv, substituting values gives this wavelength.
Q2: A) 0.061 nm - λ = 1.227/√400 = 0.061 nm.
Q3: B) Halves - λ ∝ 1/v, doubling velocity halves wavelength.

Memory Aids

De Broglie wavelength λ = h/mv (momentum inverse relation).
Electron wavelength λ = 1.227/√V nm for accelerating voltage V.
Wave-particle duality: matter behaves like waves when moving.

Concepts and Formulas

Every moving particle exhibits wave-like properties (de Broglie hypothesis).
De Broglie wavelength λ depends on particle momentum p = mv.
Formula: λ = h/p = h/mv, where h = 6.626 × 10^-34 Js.
For an electron accelerated by voltage V, λ = h/√(2meV) = 1.227/√V nm.
Wave-particle duality applies to matter similar to radiation.
De Broglie waves are also called matter waves.
Louis de Broglie received the Nobel Prize in 1929 for this discovery.

Solved Examples

Ex 1: Calculate de Broglie wavelength of an electron moving at speed 2 × 10⁶ m/s.
- Given: m_e = 9.11 × 10^-31 kg, v = 2 × 10⁶ m/s, h = 6.626 × 10^-34 Js
- λ = h/mv = 6.626 × 10^-34/(9.11 × 10^-31 × 2 × 10⁶) = 3.64 × 10^-10 m = 0.364 nm
Ex 2: Find λ for electron accelerated by V = 100 V.
- λ = 1.227/√V = 1.227/√100 = 0.1227 nm

Practice Set

Q1: What is the de Broglie wavelength of a particle with mass 2 × 10^-27 kg moving at 3 × 10³ m/s?
- A) 1.1 × 10^-34 m
- B) 1.1 × 10^-31 m
- C) 1.1 × 10^-37 m
- D) 1.1 × 10^-28 m
Q2: An electron accelerated through 400 V has de Broglie wavelength approximately:
- A) 0.061 nm
- B) 0.122 nm
- C) 0.031 nm
- D) 0.244 nm
Q3: If the velocity of a particle doubles, its de Broglie wavelength:
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: A) 1.1 × 10^-34 m - Using λ = h/mv, substituting values gives this wavelength.
Q2: A) 0.061 nm - λ = 1.227/√400 = 0.061 nm.
Q3: B) Halves - λ ∝ 1/v, doubling velocity halves wavelength.

Memory Aids

De Broglie wavelength λ = h/mv (momentum inverse relation).
Electron wavelength λ = 1.227/√V nm for accelerating voltage V.
Wave-particle duality: matter behaves like waves when moving.

Concepts and Formulas

Every moving particle exhibits wave-like properties (de Broglie hypothesis).
De Broglie wavelength \(\lambda\) depends on particle momentum \(p = mv\).
Formula: \(\lambda = \frac{h}{p} = \frac{h}{mv}\), where \(h = 6.626 \times 10^{-34} \text{ Js}\).
For an electron accelerated by voltage \(V\), \(\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.227}{\sqrt{V}} \text{ nm}\).
Wave-particle duality applies to matter similar to radiation.
De Broglie waves are also called matter waves.
Louis de Broglie received the Nobel Prize in 1929 for this discovery.

Solved Examples

Ex 1: Calculate de Broglie wavelength of an electron moving at speed \(2 \times 10^6 \text{ m/s}\).
- Given: \(m_e = 9.11 \times 10^{-31} \text{ kg}\), \(v = 2 \times 10^6 \text{ m/s}\), \(h = 6.626 \times 10^{-34} \text{ Js}\)
- \(\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2 \times 10^6} = 3.64 \times 10^{-10} \text{ m} = 0.364 \text{ nm}\)
Ex 2: Find \(\lambda\) for electron accelerated by \(V = 100 \text{ V}\).
- \(\lambda = \frac{1.227}{\sqrt{V}} = \frac{1.227}{\sqrt{100}} = 0.1227 \text{ nm}\)

Practice Set

Q1: What is the de Broglie wavelength of a particle with mass \(2 \times 10^{-27} \text{ kg}\) moving at \(3 \times 10^3 \text{ m/s}\)?
- A) \(1.1 \times 10^{-34} \text{ m}\)
- B) \(1.1 \times 10^{-31} \text{ m}\)
- C) \(1.1 \times 10^{-37} \text{ m}\)
- D) \(1.1 \times 10^{-28} \text{ m}\)
Q2: An electron accelerated through 400 V has de Broglie wavelength approximately:
- A) 0.061 nm
- B) 0.122 nm
- C) 0.031 nm
- D) 0.244 nm
Q3: If the velocity of a particle doubles, its de Broglie wavelength:
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: A) \(1.1 \times 10^{-34} \text{ m}\) - Using \(\lambda = \frac{h}{mv}\), substituting values gives this wavelength.
Q2: A) 0.061 nm - \(\lambda = \frac{1.227}{\sqrt{400}} = 0.061\) nm.
Q3: B) Halves - \(\lambda \propto \frac{1}{v}\), doubling velocity halves wavelength.

Memory Aids

De Broglie wavelength \(\lambda = \frac{h}{mv}\) (momentum inverse relation).
Electron wavelength \(\lambda = \frac{1.227}{\sqrt{V}}\) nm for accelerating voltage \(V\).
Wave-particle duality: matter behaves like waves when moving.

NEET Physics — ALL CHAPTERS

Basic Mathematics

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Units and Measurement

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Motion in a Straight Line

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Motion in a Plane

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Laws of Motion

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Work, Energy and Power

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System of Particles and Rotational Motion

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Gravitation

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Mechanical Properties of Solids

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Mechanical Properties of Fluids

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Thermal Properties of Matter

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Thermodynamics

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Kinetic Theory

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Oscillations

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Waves

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Electric Charges and Fields

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Electrostatic Potential and Capacitance

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Current Electricity

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Moving Charges and Magnetism

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Magnetism and Matter

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Electromagnetic Induction

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Alternating Current

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Electromagnetic Waves

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Ray Optics and Optical Instruments

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Wave Optics

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Dual Nature of Radiation and Matter

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Atoms

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Nuclei

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Semiconductor Electronics: Materials, Devices and Simple Circuits

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