Class 12 Physics Notes – Chapter 11: Dual Nature of Radiation and Matter
1. Electron Emission
In metals, the outer electrons (valence electrons) are loosely bound to the atoms and are free to move within the metal surface. However, they cannot leave the metal surface on their own due to the attractive pull of the positive ions.
Work Function (φ0): The minimum amount of energy required by an electron to just escape from the metal surface is called the work function of that metal. It is generally measured in electron volts (eV).
1 eV = 1.602 × 10-19 J
Electrons can be emitted from a metal surface by supplying energy through three main physical processes:
- Thermionic Emission: By heating the metal to a high temperature.
- Field Emission: By applying a very strong electric field (of the order of 108 V/m).
- Photoelectric Emission: By illuminating the metal surface with light of suitable frequency.
2. Photoelectric Effect
The phenomenon of emission of electrons from a metal surface when light of suitable high frequency falls on it is called the photoelectric effect. The emitted electrons are called photoelectrons.
Experimental Study of Photoelectric Effect
- Effect of Intensity of Light: The number of photoelectrons emitted per second (photoelectric current) is directly proportional to the intensity of incident light.
- Effect of Potential (Stopping Potential): For a given frequency of incident radiation, the minimum negative (retarding) potential V0 given to the anode for which the photoelectric current stops or becomes zero is called the cut-off or stopping potential. Maximum kinetic energy of photoelectrons is: Kmax = e V0.
- Effect of Frequency (Threshold Frequency): For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency (ν0), below which no emission of photoelectrons takes place, no matter how intense the light is.
[Place Image Here: Graph showing the variation of photoelectric current with collector plate potential for different intensities]
3. Einstein’s Photoelectric Equation
In 1905, Albert Einstein proposed that radiation energy is built up of discrete units called quanta of energy of radiation (later called photons). Each quantum has energy E = hν, where h is Planck’s constant and ν is the frequency of light.
When a photon of energy hν falls on a metal surface, a part of its energy is used to overcome the work function (φ0), and the remaining energy is given to the ejected electron as maximum kinetic energy.
Einstein’s Equation: Kmax = hν – φ0
Also written as: e V0 = hν – φ0
This equation perfectly explains why photoelectric emission is instantaneous and why a threshold frequency (ν0 = φ0 / h) exists.
4. Particle Nature of Light: The Photon
The photoelectric effect gave evidence that light in interaction with matter behaves as if it was made of quanta or packets of energy, called photons.
- Each photon has energy E = hν = hc / λ and momentum p = hν / c = h / λ.
- All photons of a particular frequency ν have the same energy and momentum, regardless of the intensity of radiation.
- Photons are electrically neutral and are not deflected by electric and magnetic fields.
5. Wave Nature of Matter
In 1924, the French physicist Louis de Broglie proposed that material particles (like electrons, protons) in motion should also display wave-like properties. The waves associated with moving material particles are called matter waves or de Broglie waves.
de Broglie Wavelength (λ): λ = h / p = h / (mv)
For an electron accelerated from rest through a potential difference of V volts, the de Broglie wavelength is given by:
λ = 1.227 / √V nm
[Place Image Here: Diagram illustrating the concept of de Broglie matter waves for a moving electron]
Multiple Choice Questions (MCQs)
1. The minimum energy required by an electron to just escape from the metal surface is called:
a) Kinetic energy
b) Work function
c) Potential energy
d) Threshold energy
Answer: (b) Work function
2. 1 electron volt (eV) is equal to:
a) 1.6 × 10-19 Joules
b) 9.1 × 10-31 Joules
c) 6.63 × 10-34 Joules
d) 3 × 108 Joules
Answer: (a) 1.6 × 10-19 Joules
3. The phenomenon of photoelectric effect was explained by Einstein using:
a) Wave theory of light
b) Particle (quantum) theory of light
c) Electromagnetic theory
d) Corpuscular theory
Answer: (b) Particle (quantum) theory of light
4. In photoelectric effect, the stopping potential depends on:
a) Intensity of incident light
b) Frequency of incident light and nature of material
c) Time of illumination
d) Area of the metal surface
Answer: (b) Frequency of incident light and nature of material
5. The momentum of a photon of wavelength λ is:
a) hλ
b) h / λ
c) λ / h
d) hc / λ
Answer: (b) h / λ
6. The de Broglie wavelength (λ) associated with a moving particle of mass m and velocity v is:
a) λ = mv / h
b) λ = h / mv
c) λ = hm / v
d) λ = h / m
Answer: (b) λ = h / mv
7. If the intensity of incident light is increased, what happens to the photoelectric current?
a) It decreases
b) It remains unchanged
c) It increases
d) It becomes zero
Answer: (c) It increases
8. Photons are deflected by:
a) Electric field only
b) Magnetic field only
c) Both electric and magnetic fields
d) Neither electric nor magnetic fields
Answer: (d) Neither electric nor magnetic fields
9. Einstein’s photoelectric equation is given by:
a) Kmax = hν + φ0
b) Kmax = hν / φ0
c) Kmax = hν – φ0
d) Kmax = φ0 – hν
Answer: (c) Kmax = hν – φ0
10. The cut-off frequency below which no photoelectric emission takes place is called:
a) Resonant frequency
b) Threshold frequency
c) Fundamental frequency
d) Natural frequency
Answer: (b) Threshold frequency
Textbook Exercise Solutions
Q 11.1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given: Accelerating voltage V = 30 kV = 30000 V.
Energy of electron E = e V = 1.6 × 10-19 × 30000 = 4.8 × 10-15 J.
(a) Maximum frequency (νmax): We know E = hνmax
νmax = E / h = (4.8 × 10-15) / (6.63 × 10-34) = 7.24 × 1018 Hz.
(b) Minimum wavelength (λmin): λmin = c / νmax
λmin = (3 × 108) / (7.24 × 1018) = 0.0414 × 10-9 m = 0.041 nm.
Q 11.2: The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
Solution:
Given: φ0 = 2.14 eV, ν = 6 × 1014 Hz.
Energy of incident photon E = hν = (6.63 × 10-34 × 6 × 1014) J = 3.978 × 10-19 J.
In eV: E = (3.978 × 10-19) / (1.6 × 10-19) = 2.486 eV.
(a) Maximum kinetic energy (Kmax) = E – φ0 = 2.486 – 2.14 = 0.34 eV (approx).
(In Joules: Kmax = 0.34 × 1.6 × 10-19 = 0.544 × 10-19 J).
(b) Stopping potential V0 = Kmax / e = 0.34 eV / e = 0.34 V.
(c) Maximum speed (vmax): 1/2 m vmax2 = Kmax
vmax = √(2 × Kmax / m) = √(2 × 0.544 × 10-19 / 9.1 × 10-31) = √(0.119 × 1012) ≈ 344 × 103 m/s = 344 km/s.
Q 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Solution:
Given: Stopping potential (cut-off voltage) V0 = 1.5 V.
Maximum kinetic energy Kmax = e V0 = e (1.5 V) = 1.5 eV.
In Joules: Kmax = 1.5 × 1.6 × 10-19 = 2.4 × 10-19 J.
Q 11.4: Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume uniform cross-section).
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Solution:
Given: λ = 632.8 nm = 632.8 × 10-9 m, Power P = 9.42 × 10-3 W.
(a) Energy of photon E = hc / λ = (6.63 × 10-34 × 3 × 108) / (632.8 × 10-9) = 3.14 × 10-19 J.
Momentum p = h / λ = (6.63 × 10-34) / (632.8 × 10-9) = 1.05 × 10-27 kg m/s.
(b) Number of photons per second N = P / E = (9.42 × 10-3) / (3.14 × 10-19) = 3 × 1016 photons/s.
(c) Momentum of Hydrogen atom p = m × v. Mass of H-atom m = 1.67 × 10-27 kg.
v = p / m = (1.05 × 10-27) / (1.67 × 10-27) = 0.63 m/s.
Q 11.5: In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.
Solution:
We know from Einstein’s photoelectric equation: V0 = (h/e)ν – (φ0/e).
The slope of the V0 vs ν graph is equal to h/e.
Slope = h / e = 4.12 × 10-15 V s.
Planck’s constant h = e × Slope = 1.6 × 10-19 × 4.12 × 10-15 = 6.59 × 10-34 J s.
Q 11.6: The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution:
Given: Threshold frequency ν0 = 3.3 × 1014 Hz, Incident frequency ν = 8.2 × 1014 Hz.
According to Einstein’s equation: e V0 = h(ν – ν0)
V0 = (h/e) × (ν – ν0) = [(6.63 × 10-34) / (1.6 × 10-19)] × (8.2 – 3.3) × 1014
V0 = (4.14 × 10-15) × (4.9 × 1014) = 2.0286 V ≈ 2.0 V.
Q 11.7: The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Solution:
Given: Work function φ0 = 4.2 eV, Wavelength λ = 330 nm = 330 × 10-9 m.
Energy of incident photon E = hc / λ
E = (6.63 × 10-34 × 3 × 108) / (330 × 10-9) = 6.027 × 10-19 J.
In eV: E = (6.027 × 10-19) / (1.6 × 10-19) ≈ 3.76 eV.
Since the incident energy E (3.76 eV) is less than the work function φ0 (4.2 eV), No, photoelectric emission will not take place.