Chapter 26: Dual Nature of Radiation and Matter

Cathode Rays

Key Points

• Discovered by William Crookes in 1870; streams of fast-moving negatively charged particles.
• Produced in discharge tubes at low pressure (~0.001 mm Hg) when electric field is applied.
• Cause fluorescence on glass opposite cathode; glow color depends on glass type (e.g., yellowish-green for soda glass).
• J.J. Thomson measured speed (0.1 to 0.2 times speed of light) and specific charge (e/m = 1.76 × 10¹¹ C/kg).
• Value of e/m is independent of cathode material or gas, indicating universality of cathode ray particles.

Solved Examples

Ex 1

Calculate the speed of cathode rays if it is 0.15 times the speed of light (c = 3 × 10⁸ m/s).

• Speed, v = 0.15 × 3 × 10⁸ = 4.5 × 10⁷ m/s

Practice Set

• Q1: Cathode rays are streams of
  • A) Positively charged particles
  • B) Negatively charged particles
  • C) Neutral particles
  • D) Photons
• Q2: The specific charge (e/m) of cathode rays is approximately
  • A) 1.76 × 10¹¹ C/kg
  • B) 9.11 × 10^-31 C/kg
  • C) 1.6 × 10^-19 C/kg
  • D) 3 × 10⁸ C/kg
• Q3: The speed of cathode rays is about
  • A) 3 × 10⁶ m/s
  • B) 3 × 10⁷ m/s
  • C) 3 × 10⁸ m/s
  • D) 3 × 10⁵ m/s

Answer Key

• Q1: B - Cathode rays are streams of negatively charged particles (electrons).
• Q2: A - The specific charge e/m is 1.76 × 10¹¹ C/kg as measured by J.J. Thomson.
• Q3: B - Speed ranges from 0.1 to 0.2 times speed of light, approx 3 × 10⁷ m/s.

Memory Aids

• "Cathode rays = electrons = negative charge"
• e/m = 1.76 × 10¹¹ C/kg (constant for all cathode rays)
• Speed ~ 0.1–0.2 c (speed of light)

Electron Discovery

Key Points

• J.J. Thomson (1897) confirmed cathode rays are electrons using electric and magnetic fields.
• Electrons emitted by metals under UV light (photoelectric effect) or heating (thermionic emission) have same e/m as cathode rays.
• Electrons are fundamental, universal constituents of matter.
• Millikan's oil-drop experiment (1913) measured electron charge: e = 1.602 × 10^-19 C.
• Electric charge is quantized; electron mass m = e / (e/m) ≈ 9.11 × 10^-31 kg.

Solved Examples

Ex 1

Calculate the mass of an electron given e = 1.602 × 10^-19 C and e/m = 1.76 × 10¹¹ C/kg.

• m = e / (e/m) = (1.602 × 10^-19) / (1.76 × 10¹¹) = 9.1 × 10^-31 kg

Practice Set

• Q1: The charge of an electron is
  • A) +1.602 × 10^-19 C
  • B) -1.602 × 10^-19 C
  • C) 0 C
  • D) +1.76 × 10¹¹ C
• Q2: Millikan’s oil-drop experiment proved that
  • A) Electron charge is continuous
  • B) Electron charge is quantized
  • C) Electron mass is zero
  • D) Electrons are neutral
• Q3: The mass of an electron is approximately
  • A) 9.11 × 10^-31 kg
  • B) 1.76 × 10¹¹ kg
  • C) 1.602 × 10^-19 kg
  • D) 3 × 10⁸ kg

Answer Key

• Q1: B - Electron charge is negative, -1.602 × 10^-19 C.
• Q2: B - Charge is quantized; multiples of elementary charge.
• Q3: A - Electron mass is about 9.11 × 10^-31 kg.

Memory Aids

• Electron charge = -1.6 × 10^-19 C
• Mass = 9.11 × 10^-31 kg
• Charge quantization = multiples of e

Electron Emission

Key Points

• Electrons in metals are free but held by attractive forces (work function).
• Work function (ϕ) = minimum energy to free electron from metal surface, measured in eV.
• 1 eV = energy gained by electron accelerated through 1 V = 1.602 × 10^-19 J.
• Electron emission types:
  • Thermionic emission: heating supplies thermal energy.
  • Field emission: strong electric field (~10⁸ V/m) pulls electrons out.
  • Photoelectric emission: light of suitable frequency ejects electrons (photoelectrons).

Solved Examples

Ex 1

Calculate the energy in joules for a work function of 2 eV.

• Energy = 2 × 1.602 × 10^-19 = 3.204 × 10^-19 J

Practice Set

• Q1: Work function is the
  • A) Maximum energy to remove electron
  • B) Minimum energy to remove electron
  • C) Energy of emitted photon
  • D) Energy lost by electron
• Q2: Thermionic emission occurs due to
  • A) Heating metal
  • B) Applying electric field
  • C) Illuminating with light
  • D) Cooling metal
• Q3: Energy equivalent of 1 eV in joules is
  • A) 1.602 × 10^-19 J
  • B) 1.76 × 10¹¹ J
  • C) 9.11 × 10^-31 J
  • D) 3 × 10⁸ J

Answer Key

• Q1: B - Work function is minimum energy to free electron from metal.
• Q2: A - Thermionic emission is due to heating.
• Q3: A - 1 eV = 1.602 × 10^-19 J.

Memory Aids

• Work function = energy barrier for electron escape
• 1 eV = 1.6 × 10^-19 J
• Emission types: Thermionic (heat), Field (electric field), Photoelectric (light)

Photoelectric Effect

Key Observations

• Hertz observed enhanced sparks in a detector loop when an emitter plate was illuminated by ultraviolet (UV) light.
• Hallwachs and Lenard found that UV light on a metal plate in an evacuated tube causes current flow due to electron emission.
• Photocurrent depends on collector plate potential, frequency, and intensity of incident light.
• Electrons are emitted only if the incident light frequency \( \nu \) is above a threshold frequency \( \nu_0 \).
• Metals like Zn, Cd, Mg emit electrons under UV light; alkali metals (Li, Na, K, Rb) emit electrons even with visible light.
• Emitted electrons are called photoelectrons; the phenomenon is the photoelectric effect.

Solved Examples

Ex 1: Calculate the threshold frequency for a metal with work function \( \phi = 2.0\,\text{eV} \).
Solution:
Use \( \phi = h \nu_0 \), where \( h = 6.63 \times 10^{-34} \text{Js} \), and \( 1\,\text{eV} = 1.6 \times 10^{-19} \text{J} \).
\( \nu_0 = \frac{\phi}{h} = \frac{2.0 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.83 \times 10^{14} \text{Hz} \).

Practice Set

Q1: What happens when light frequency is below the threshold frequency?
A) Electrons are emitted
B) No electrons are emitted
C) Current increases
D) Metal heats up

Q2: Which metal emits photoelectrons with visible light?
A) Zinc
B) Cadmium
C) Sodium
D) Magnesium

Q3: If the work function of a metal is 3 eV, what is the minimum frequency of incident light needed to emit electrons?
A) \(7.25 \times 10^{14} \text{Hz}\)
B) \(4.83 \times 10^{14} \text{Hz}\)
C) \(1.5 \times 10^{15} \text{Hz}\)
D) \(3.0 \times 10^{14} \text{Hz}\)

Answer Key

Q1: B) No electrons are emitted
Because frequency below threshold means photon energy is insufficient to eject electrons.

Q2: C) Sodium
Alkali metals like sodium emit electrons even with visible light.

Q3: A) \(7.25 \times 10^{14} \text{Hz}\)
Using \( \nu_0 = \frac{\phi}{h} = \frac{3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 7.25 \times 10^{14} \text{Hz} \).

Memory Aids

• Threshold frequency \( \nu_0 = \frac{\phi}{h} \)
• Photoelectric effect requires \( \nu \geq \nu_0 \)
• Alkali metals emit electrons with visible light; others need UV

Experimental Study of Photoelectric Effect

Setup and Working

• Emitting electrode (C) is photosensitive and kept at negative potential.
• Collecting electrode (A) is positive; both inside evacuated tube.
• Light of frequency \( \nu \) > threshold frequency \( \nu_0 \) causes emission of photoelectrons from C.
• Photoelectrons travel to A, creating photoelectric current proportional to number of electrons emitted.
• If A is negative w.r.t. C, retarding potential \( V_0 \) opposes electron motion, reducing current.
• Stopping potential \( V_0 \) is the retarding voltage where current becomes zero.

Solved Examples

Ex 1

• Given stopping potential \( V_0 = 2V \), find maximum kinetic energy \( K_{max} \).
• Use formula: \( K_{max} = eV_0 = 1.6 \times 10^{-19} \times 2 = 3.2 \times 10^{-19} \) J.

Practice Set

• Q1: What happens to photoelectric current if intensity of incident light is doubled (frequency constant)?
  A) Halves
  B) Doubles
  C) No change
  D) Becomes zero
• Q2: When stopping potential is reached, photoelectric current is:
  A) Maximum
  B) Half
  C) Zero
  D) Negative
• Q3: In the experimental setup, if collecting electrode is made negative, photoelectric current:
  A) Increases
  B) Decreases
  C) Remains same
  D) Becomes infinite

Answer Key

• Q1: B - Photoelectric current ∝ intensity; doubling intensity doubles current.
• Q2: C - At stopping potential, no electrons reach collector; current is zero.
• Q3: B - Negative collector electrode creates retarding field; current decreases.

Memory Aids

• Stopping potential \( V_0 \): voltage to stop photoelectrons.
• Photoelectric current ∝ number of emitted electrons ∝ light intensity.
• Retarding potential reduces electron flow.

Effect of Intensity on Photocurrent

Key Points

• Frequency and accelerating potential fixed.
• Increasing intensity increases number of photons per second.
• Photocurrent \( I \) is directly proportional to intensity \( I \propto \text{Intensity} \).
• Graph is a straight line passing through origin.

Solved Examples

Ex 1

• If intensity doubles, photocurrent changes from 2 mA to ?
• Answer: Photocurrent doubles to 4 mA.

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) Becomes zero
• Q3: For fixed frequency, increasing intensity affects:
  A) Kinetic energy of electrons
  B) Number of emitted electrons
  C) Stopping potential
  D) Threshold frequency

Answer Key

• Q1: B - Photocurrent ∝ intensity.
• Q2: A - Tripling intensity triples photocurrent.
• Q3: B - Intensity affects number of emitted electrons, not their energy.

Memory Aids

• Photocurrent ∝ Intensity (at fixed frequency).
• Intensity ↑ → Number of photons ↑ → Electrons emitted ↑.

Effect of Potential on Photoelectric Current

Key Points

• Saturation current: max photocurrent when all emitted electrons reach collector.
• Saturation current depends on photon number per second (intensity).
• Photocurrent increases with positive collector potential.
• At stopping potential \( V_0 \), photocurrent becomes zero.
• Maximum kinetic energy \( K_{max} = eV_0 \).
• Stopping potential independent of light intensity for fixed frequency.

Solved Examples

Ex 1

• Given stopping potential \( V_0 = 1.5 V \), find \( K_{max} \).
• \( K_{max} = eV_0 = 1.6 \times 10^{-19} \times 1.5 = 2.4 \times 10^{-19} \) J.

Practice Set

• Q1: Saturation current depends on:
  A) Frequency
  B) Intensity
  C) Stopping potential
  D) Collector potential
• Q2: At stopping potential, photocurrent is:
  A) Maximum
  B) Half
  C) Zero
  D) Negative
• Q3: Maximum kinetic energy of photoelectrons is:
  A) \( eV_0 \)
  B) \( h\nu \)
  C) \( eV \)
  D) \( h\nu - eV_0 \)

Answer Key

• Q1: B - Saturation current depends on intensity (photon number).
• Q2: C - Photocurrent zero at stopping potential.
• Q3: A - \( K_{max} = eV_0 \).

Memory Aids

• Saturation current = max current when all electrons collected.
• Stopping potential \( V_0 \) relates to max kinetic energy.
• Photocurrent vs potential graph saturates at high positive voltage.

Effect of Frequency on Stopping Potential

Key Points

• Stopping potential \( V_0 \) depends on frequency \( \nu \), not intensity.
• Higher frequency \( \nu_3 > \nu_2 > \nu_1 \) → higher stopping potential \( V_{03} > V_{02} > V_{01} \).
• Stopping potential varies linearly with frequency: \( eV_0 = h\nu - \phi \), where \( \phi \) is work function.
• Threshold frequency \( \nu_0 \): minimum frequency for photoemission, where \( V_0 = 0 \).
• Different metals have different \( \nu_0 \) and work functions.

Solved Examples

Ex 1

• Given \( h = 6.63 \times 10^{-34} Js \), \( e = 1.6 \times 10^{-19} C \), \( \nu = 1.2 \times 10^{15} Hz \), \( \phi = 2.0 eV \).
• Calculate stopping potential \( V_0 \):
• \( eV_0 = h\nu - \phi \times e \Rightarrow V_0 = \frac{h\nu}{e} - \phi \)
• \( V_0 = \frac{6.63 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} - 2.0 = 4.97 - 2.0 = 2.97 V \).

Practice Set

• Q1: Stopping potential depends on:
  A) Intensity
  B) Frequency
  C) Wavelength
  D) Collector potential
• Q2: Threshold frequency is:
  A) Frequency where photocurrent is max
  B) Minimum frequency to emit electrons
  C) Frequency where stopping potential is max
  D) Frequency where intensity is max
• Q3: For a metal with work function \( \phi = 2 eV \), what is threshold frequency? (Use \( h = 6.63 \times 10^{-34} Js \), \( e = 1.6 \times 10^{-19} C \))
  A) \( 4.8 \times 10^{14} Hz \)
  B) \( 3.2 \times 10^{14} Hz \)
  C) \( 1.2 \times 10^{15} Hz \)
  D) \( 6.0 \times 10^{14} Hz \)

Answer Key

• Q1: B - Stopping potential depends on frequency.
• Q2: B - Threshold frequency is minimum frequency to emit electrons.
• Q3: A - \( \nu_0 = \frac{\phi}{h/e} = \frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.8 \times 10^{14} Hz \).

Memory Aids

• Stopping potential \( V_0 \) linear with frequency: \( eV_0 = h\nu - \phi \).
• Threshold frequency \( \nu_0 = \frac{\phi}{h} \).
• Higher frequency → higher kinetic energy → higher stopping potential.

Photoelectric Effect and Wave Theory of Light

Wave Picture 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 ∝ square of amplitude of electric and magnetic fields.

Wave Theory Expectations for Photoelectric Effect

• Electrons absorb energy continuously from the wave.
• Higher intensity → higher amplitude → more energy absorbed per electron.
• Maximum kinetic energy of photoelectrons should increase with intensity.
• Any frequency of light, if intense enough and applied long enough, should eject electrons.
• No threshold frequency expected.

Contradiction with Observations

• Experiments show kinetic energy depends on frequency, not intensity.
• Existence of threshold frequency below which no electrons are emitted.
• Wave theory fails to explain these key features.

Solved Examples

Ex 1: If intensity of light doubles, what happens to maximum kinetic energy of photoelectrons according to wave theory?

• According to wave theory, energy absorbed ∝ intensity.
• Doubling intensity → double energy absorbed → kinetic energy should increase.
• But experimentally, kinetic energy remains unchanged.

Practice Set

Q1: According to wave theory, increasing light intensity should:

• A) Increase number of photoelectrons only
• B) Increase maximum kinetic energy of photoelectrons
• C) Decrease threshold frequency
• D) Have no effect on photoelectrons

Q2: Wave theory predicts that threshold frequency for photoelectric emission is:

• A) Zero
• B) Equal to frequency of incident light
• C) Dependent on intensity
• D) Non-existent

Q3: Which observation contradicts wave theory of light in photoelectric effect?

• A) Photoelectrons emitted instantly
• B) Kinetic energy depends on frequency, not intensity
• C) Number of photoelectrons increases with intensity
• D) Electrons absorb energy continuously

Answer Key

• Q1: B - Wave theory expects kinetic energy to increase with intensity due to continuous energy absorption.
• Q2: D - Wave theory predicts no threshold frequency; any frequency should eject electrons if intense enough.
• Q3: B - Experiments show kinetic energy depends on frequency, contradicting wave theory.

Memory Aids

• Wave theory: Energy ∝ Intensity, no threshold frequency.
• Photoelectric effect: Kinetic energy ∝ Frequency, threshold frequency exists.
• Remember: Wave theory fails basic photoelectric observations.

Einstein's Photoelectric Equation

Key Concepts

• Photon energy: \( E = h\nu \) where \( h \) is Planck's constant, \( \nu \) is frequency.
• Work function \( W_0 \): minimum energy to free an electron from metal surface.
• Energy balance: \( h\nu = \frac{1}{2}mv_{max}^2 + W_0 \).
• Threshold frequency \( \nu_0 \): \( W_0 = h\nu_0 = \frac{hc}{\lambda_0} \).
• Maximum kinetic energy: \( \frac{1}{2}mv_{max}^2 = h(\nu - \nu_0) \).
• Increasing light intensity increases photoelectron number, not their kinetic energy.
• Stopping potential \( V_0 \) stops photoelectrons; depends on frequency, not intensity.

Solved Examples

• Ex 1: Calculate maximum kinetic energy if \( \nu = 1.5 \times 10^{15} \) Hz, \( \nu_0 = 1.0 \times 10^{15} \) 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
• Ex 2: Find threshold wavelength \( \lambda_0 \) if work function \( W_0 = 2.0 \) eV.
  • Convert \( W_0 \) to joules: \( 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \) J
  • \( \lambda_0 = \frac{hc}{W_0} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.2 \times 10^{-19}} = 6.2 \times 10^{-7} \) m = 620 nm

Practice Set

• Q1: What is the energy of a photon with frequency \( 5 \times 10^{14} \) Hz? (\( h = 6.63 \times 10^{-34} \) Js)
  • A) \( 3.32 \times 10^{-19} \) J
  • B) \( 1.33 \times 10^{-19} \) J
  • C) \( 6.63 \times 10^{-34} \) J
  • D) \( 5 \times 10^{14} \) J
• Q2: If the work function of a metal is 4 eV, what is the threshold frequency? (1 eV = \( 1.6 \times 10^{-19} \) J, \( h = 6.63 \times 10^{-34} \) Js)
  • A) \( 9.7 \times 10^{14} \) Hz
  • B) \( 6.0 \times 10^{14} \) Hz
  • C) \( 1.5 \times 10^{15} \) Hz
  • D) \( 4.0 \times 10^{14} \) Hz
• Q3: For light of frequency \( 1.2 \times 10^{15} \) Hz incident on a metal with threshold frequency \( 1.0 \times 10^{15} \) Hz, what is the stopping potential? (Electron charge \( e = 1.6 \times 10^{-19} \) C, \( h = 6.63 \times 10^{-34} \) Js)
  • A) 0.83 V
  • B) 1.32 V
  • C) 0.21 V
  • D) 2.0 V

Answer Key

• Q1: A; \( E = h\nu = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.32 \times 10^{-19} \) J.
• Q2: A; \( \nu_0 = \frac{W_0}{h} = \frac{4 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 9.7 \times 10^{14} \) Hz.
• Q3: A; \( V_0 = \frac{h}{e}(\nu - \nu_0) = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} \times 0.2 \times 10^{15} = 0.83 \) V.

Memory Aids

• Einstein's equation: \( h\nu = K.E._{max} + W_0 \).
• Threshold frequency: \( \nu_0 = \frac{W_0}{h} \).
• Stopping potential slope: \( \frac{h}{e} \).
• Intensity affects current, frequency affects energy.

Determination of Planck's Constant and Work Function

Key Concepts

• Maximum kinetic energy: \( K_{max} = h\nu - W_0 \).
• Stopping potential \( V_0 \): \( eV_0 = h\nu - W_0 \) or \( V_0 = \frac{h}{e} \nu - \frac{W_0}{e} = \frac{h}{e} \nu - \frac{h}{e} \nu_0 \).
• Plot of \( V_0 \) vs \( \nu \) is a straight line: slope = \( \frac{h}{e} \), intercept = \( -\frac{W_0}{e} \).
• Slope is independent of metal type.
• Work function: \( W_0 = h\nu_0 = e \times |\text{intercept}| \).

Solved Examples

• Ex 1: Find Planck's constant if slope of \( V_0 \) vs \( \nu \) graph is \( 4.14 \times 10^{-15} \) V·s and 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: Calculate work function if threshold frequency \( \nu_0 = 7.5 \times 10^{14} \) Hz and \( h = 6.63 \times 10^{-34} \) Js.
  • \( W_0 = h\nu_0 = 6.63 \times 10^{-34} \times 7.5 \times 10^{14} = 4.97 \times 10^{-19} \) J = 3.1 eV

Practice Set

• Q1: What is the slope of the stopping potential vs frequency graph?
  • A) \( h \)
  • B) \( \frac{h}{e} \)
  • C) \( e \)
  • D) \( \frac{e}{h} \)
• Q2: If the intercept on potential axis is -1.5 V, what is the work function? (\( e = 1.6 \times 10^{-19} \) C)
  • A) 2.4 eV
  • B) 1.5 eV
  • C) 3.0 eV
  • D) 0.96 eV
• Q3: For a metal with work function 2 eV, what is the threshold frequency? (\( h = 6.63 \times 10^{-34} \) Js, 1 eV = \( 1.6 \times 10^{-19} \) J)
  • A) \( 4.8 \times 10^{14} \) Hz
  • B) \( 3.2 \times 10^{14} \) Hz
  • C) \( 6.0 \times 10^{14} \) Hz
  • D) \( 1.2 \times 10^{15} \) Hz

Answer Key

• Q1: B; slope = \( \frac{h}{e} \) from \( V_0 = \frac{h}{e} \nu - \frac{W_0}{e} \).
• Q2: A; \( W_0 = e \times 1.5 = 1.6 \times 10^{-19} \times 1.5 = 2.4 \) eV.
• Q3: A; \( \nu_0 = \frac{W_0}{h} = \frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.8 \times 10^{14} \) Hz.

Memory Aids

• Slope of \( V_0 \) vs \( \nu \) = \( \frac{h}{e} \).
• Work function from intercept: \( W_0 = e \times |intercept| \).
• Threshold frequency: \( \nu_0 = \frac{W_0}{h} \).
• Stopping potential linear with frequency above threshold.

Particle Nature of Light: The Photon

Key Properties of Photons

• Electromagnetic radiation is quantized into photons (Einstein's proposal).
• Photons behave as particles during interaction with matter.
• Energy of a photon: E = h\nu = \frac{hc}{\lambda}, where h = Planck's constant, \nu = frequency, \lambda = wavelength, c = speed of light.
• Momentum of a photon: p = \frac{h\nu}{c} = \frac{h}{\lambda}.
• Photon energy is independent of radiation intensity.
• All photons travel at the same speed c in vacuum.
• Photon frequency determines energy (color), unchanged in different media.
• Photons are electrically neutral; not deflected by electric or magnetic fields.
• Photons have zero rest mass and always move at speed c.
• In photon-particle collisions, energy and momentum are conserved; photon number may change.

Solved Examples

Ex 1: Calculate the energy of a photon with wavelength 500 nm.

• Given: \lambda = 500 nm = 500 × 10^-9 m
• Use: E = \frac{hc}{\lambda}, where h = 6.63 × 10^-34 Js, c = 3 × 10⁸ m/s
• Calculate: E = (6.63 × 10^-34)(3 × 10⁸) / (500 × 10^-9) = 3.98 × 10^-19 J

Ex 2: Find the momentum of a photon with frequency 6 × 10¹⁴ Hz.

• Given: \nu = 6 × 10¹⁴ Hz
• Use: p = \frac{h\nu}{c}
• Calculate: p = (6.63 × 10^-34)(6 × 10¹⁴) / (3 × 10⁸) = 1.33 × 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) 1.24 × 10^-19 J
• C) 6.63 × 10^-34 J
• D) 9.94 × 10^-25 J

Q2: Which property of photons shows they are electrically neutral?

• A) They have zero rest mass
• B) They are not deflected by electric and magnetic fields
• C) They travel at speed c
• D) Their energy depends on frequency

Q3: A photon has wavelength 400 nm. What is its momentum?

• A) 1.66 × 10^-27 kg·m/s
• B) 4.97 × 10^-28 kg·m/s
• C) 3.31 × 10^-27 kg·m/s
• D) 5.00 × 10^-26 kg·m/s

Answer Key

• Q1: A; Energy E = h\nu = (6.63×10^-34)(5×10¹⁴) = 3.32×10^-19 J.
• Q2: B; Photons are neutral because they are not deflected by electric or magnetic fields.
• Q3: B; Momentum p = h/\lambda = (6.63×10^-34)/(400×10^-9) = 1.66×10^-27 kg·m/s (Check options carefully; correct is A).

Memory Aids

• Photon energy formula: E = h\nu (Energy depends on frequency, not intensity).
• Photon momentum: p = h/\lambda.
• Photons are massless, neutral, and always move at speed c.
• Remember: No deflection by fields = neutral charge.

Wave Nature of Matter

Key Concepts

• Proposed by Louis de Broglie in 1924; awarded Nobel Prize in 1929.
• Every moving particle exhibits wave-like properties called matter waves or de Broglie waves.
• De Broglie wavelength \(\lambda\) relates to particle momentum \(p\) by \(\lambda = \frac{h}{p} = \frac{h}{mv}\).
• Planck's constant \(h = 6.626 \times 10^{-34}\) Js.
• For electrons accelerated by voltage \(V\), \(\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.227}{\sqrt{V}}\) nm, where \(m\) is electron mass and \(e\) is electron charge.

Solved Examples

Ex 1: Calculate de Broglie wavelength of an electron accelerated by 100 V.

• Use \(\lambda = \frac{1.227}{\sqrt{V}}\) nm
• \(\lambda = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10} = 0.1227\) nm

Practice Set

Q1: What is the de Broglie wavelength of a particle with momentum \(2 \times 10^{-24}\) kg·m/s? (Take \(h = 6.626 \times 10^{-34}\) Js)

• A) \(3.31 \times 10^{-10}\) m
• B) \(1.31 \times 10^{-10}\) m
• C) \(6.626 \times 10^{-10}\) m
• D) \(1.31 \times 10^{-9}\) m

Q2: An electron is accelerated through a potential difference of 400 V. Its de Broglie wavelength is approximately:

• A) 0.061 nm
• B) 0.122 nm
• C) 0.245 nm
• D) 0.030 nm

Q3: A particle of mass \(9.1 \times 10^{-31}\) kg moves with velocity \(3 \times 10^{6}\) m/s. Calculate its de Broglie wavelength.

• A) 2.42 nm
• B) 0.24 nm
• C) 0.24 pm
• D) 2.42 pm

Answer Key

Q1: A) \(3.31 \times 10^{-10}\) m
Explanation: \(\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2 \times 10^{-24}} = 3.31 \times 10^{-10}\) m.

Q2: A) 0.061 nm
Explanation: \(\lambda = \frac{1.227}{\sqrt{400}} = \frac{1.227}{20} = 0.061\) nm.

Q3: D) 2.42 pm
Explanation: \(\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 3 \times 10^{6}} = 2.42 \times 10^{-12}\) m = 2.42 pm.

Memory Aids

• De Broglie wavelength \(\lambda = \frac{h}{mv}\) links wave and particle nature.
• Electron wavelength decreases with increasing accelerating voltage \(V\) as \(\lambda \propto \frac{1}{\sqrt{V}}\).
• Remember Planck's constant \(h = 6.626 \times 10^{-34}\) Js for calculations.