CLASS 12 PHYSICS CHAPTER 11 – DUAL NATURE OF RADIATION AND MATTER
NCERT CLASS 12 PHYSICS
CHAPTER 11 – DUAL NATURE OF RADIATION AND MATTER
1. IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
Q1. What is meant by the dual nature of radiation?
Answer: The dual nature of radiation means that light exhibits both wave-like and particle-like properties.
Wave nature: Light shows interference, diffraction and polarisation.
Particle nature: Light behaves as a stream of particles called photons.
Q2. What is meant by the dual nature of matter?
Answer: The dual nature of matter means that material particles such as electrons, protons and neutrons exhibit both particle-like and wave-like properties.
The wave nature of moving particles is described by de Broglie's hypothesis.
Q3. What is electron emission?
Answer: Electron emission is the process by which electrons are released from the surface of a material when sufficient energy is supplied to overcome the attractive forces holding them inside the material.
Q4. What is work function?
Answer: Work function is the minimum energy required to remove an electron from the surface of a metal.
It is represented by φ₀.
SI unit: Joule (J)
It is commonly expressed in electron volts (eV).
Q5. What is one electron volt?
Answer: One electron volt is the energy gained by an electron when it moves through a potential difference of one volt.
1 eV = 1.6 × 10⁻¹⁹ J
Q6. What are the different methods of electron emission?
Answer: The main methods of electron emission are:
Thermionic emission: Electrons are emitted when a metal is heated to a sufficiently high temperature.
Field emission: Electrons are emitted from a metal surface under a very strong electric field.
Photoelectric emission: Electrons are emitted from a metal surface when light of suitable frequency falls on it.
Secondary emission: Electrons are emitted when energetic particles strike a material surface.
Q7. What is the photoelectric effect?
Answer: The photoelectric effect is the phenomenon of emission of electrons from a metal surface when electromagnetic radiation of suitable frequency falls on it.
The emitted electrons are called photoelectrons.
Q8. What is a photoelectron?
Answer: An electron emitted from a material surface due to the absorption of incident electromagnetic radiation is called a photoelectron.
Q9. What is photoelectric current?
Answer: The electric current produced due to the flow of photoelectrons from an illuminated surface is called photoelectric current.
Q10. What is threshold frequency?
Answer: Threshold frequency is the minimum frequency of incident radiation required to produce photoelectric emission from a particular metal surface.
It is represented by ν₀.
If the frequency of incident light is below the threshold frequency, photoelectric emission does not take place, regardless of the intensity of the light in the ordinary photoelectric effect.
Q11. What is threshold wavelength?
Answer: Threshold wavelength is the maximum wavelength of incident radiation that can just produce photoelectric emission from a particular metal.
It is represented by λ₀.
The relation is:
λ₀ = c/ν₀
Q12. What is stopping potential?
Answer: Stopping potential is the minimum negative potential applied to the collector relative to the emitter that is sufficient to stop even the most energetic photoelectrons.
It is represented by V₀.
Q13. What is saturation current?
Answer: Saturation current is the maximum photoelectric current obtained when all the photoelectrons emitted by the illuminated surface are collected.
Increasing the positive potential beyond the saturation region does not significantly increase the photoelectric current.
Q14. State the laws of the photoelectric effect.
Answer:
The main laws of the photoelectric effect are:
For a given metal, photoelectric emission takes place only when the frequency of incident light is greater than or equal to its threshold frequency.
For a frequency above the threshold frequency, the photoelectric current is directly proportional to the intensity of incident light, provided other experimental conditions remain unchanged.
The maximum kinetic energy of photoelectrons depends on the frequency of incident radiation and not on its intensity.
Photoelectric emission occurs practically instantaneously when suitable radiation falls on the metal surface.
For a given frequency above the threshold frequency, the stopping potential is independent of the intensity of incident light.
Q15. How does the intensity of incident light affect photoelectric current?
Answer: When the frequency of incident light is above the threshold frequency, increasing the intensity increases the number of photons falling on the surface per second.
This results in the emission of more photoelectrons per second and hence increases the photoelectric current.
Q16. How does the frequency of incident light affect the maximum kinetic energy of photoelectrons?
Answer: The maximum kinetic energy of photoelectrons increases with the frequency of incident light, provided the frequency is above the threshold frequency.
According to Einstein's photoelectric equation:
Kmax = hν − φ₀
Q17. Does increasing the intensity of light increase the maximum kinetic energy of photoelectrons?
Answer: No. Increasing the intensity of light does not increase the maximum kinetic energy of photoelectrons if the frequency remains unchanged.
It increases the number of emitted photoelectrons and hence the photoelectric current.
Q18. Why does photoelectric emission not occur below the threshold frequency?
Answer: Below the threshold frequency, the energy of an individual photon is less than the work function of the metal.
Therefore, an electron cannot acquire sufficient energy from a single photon to escape from the metal surface.
Q19. What is a photon?
Answer: A photon is a packet or quantum of electromagnetic radiation.
The energy of a photon is:
E = hν
Where:
E = Energy of photon
h = Planck's constant
ν = Frequency of radiation
Q20. Write the energy of a photon in terms of wavelength.
Answer:
We know:
E = hν
Also:
ν = c/λ
Therefore:
E = hc/λ
This shows that photon energy is inversely proportional to wavelength.
Q21. What is the momentum of a photon?
Answer: A photon has momentum even though its rest mass is zero.
The momentum of a photon is:
p = hν/c
Since c = νλ:
p = h/λ
Q22. What is Einstein's photoelectric equation?
Answer: Einstein's photoelectric equation is based on the conservation of energy during photoelectric emission.
It is:
hν = φ₀ + Kmax
Therefore:
Kmax = hν − φ₀
Where:
hν = Energy of incident photon
φ₀ = Work function of the metal
Kmax = Maximum kinetic energy of the emitted photoelectron
Q23. Write Einstein's photoelectric equation in terms of stopping potential.
Answer:
We know:
Kmax = eV₀
Using Einstein's equation:
eV₀ = hν − φ₀
Since φ₀ = hν₀:
eV₀ = h(ν − ν₀)
Q24. What is the relation between work function and threshold frequency?
Answer:
At threshold frequency, the emitted electron has zero maximum kinetic energy.
Therefore:
φ₀ = hν₀
Where:
φ₀ = Work function
h = Planck's constant
ν₀ = Threshold frequency
Q25. What is de Broglie's hypothesis?
Answer: According to de Broglie's hypothesis, every moving material particle is associated with a wave called a matter wave.
The wavelength associated with a moving particle is:
λ = h/p
For a non-relativistic particle:
λ = h/mv
Where:
h = Planck's constant
p = Momentum
m = Mass of particle
v = Velocity of particle
Q26. What is de Broglie wavelength?
Answer: The wavelength associated with a moving material particle is called its de Broglie wavelength.
Formula:
λ = h/mv
Q27. Why is the wave nature of ordinary objects not observed?
Answer: The de Broglie wavelength of ordinary macroscopic objects is extremely small because their masses and momenta are large.
Therefore, their wave nature cannot be observed in ordinary situations.
Q28. Why can the wave nature of electrons be observed?
Answer: Electrons have very small mass. When they move at suitable speeds, their de Broglie wavelength can be comparable to the spacing between atoms in a crystal.
Therefore, their wave nature can be demonstrated through diffraction.
Q29. What is the Davisson-Germer experiment?
Answer: The Davisson-Germer experiment demonstrated the wave nature of electrons.
In this experiment, electrons were directed towards a nickel crystal. The scattered electrons showed a diffraction pattern.
This result confirmed de Broglie's hypothesis that moving particles have wave-like properties.
Q30. What is the relation between the de Broglie wavelength and the accelerating potential of an electron?
Answer:
When an electron is accelerated through a potential difference V, it gains kinetic energy:
eV = p²/2m
Therefore:
p = √(2meV)
Using de Broglie's relation:
λ = h/p
Hence:
λ = h/√(2meV)
For an electron:
λ ≈ 1.227/√V nm
Here V is in volts.
2. NCERT EXERCISE QUESTIONS AND ANSWERS
Question 11.1
Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given:
Potential difference:
V = 30 kV
V = 30 × 10³ V
Charge of electron:
e = 1.6 × 10⁻¹⁹ C
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
Speed of light:
c = 3 × 10⁸ m/s
The maximum energy of an X-ray photon is equal to the kinetic energy gained by an electron.
Therefore:
hνmax = eV
(a) Maximum frequency
νmax = eV/h
Substituting the values:
νmax = (1.6 × 10⁻¹⁹ × 30 × 10³)/(6.63 × 10⁻³⁴)
νmax = 7.24 × 10¹⁸ Hz
(b) Minimum wavelength
We know:
λmin = c/νmax
λmin = (3 × 10⁸)/(7.24 × 10¹⁸)
λmin = 4.14 × 10⁻¹¹ m
Final Answer:
(a) Maximum frequency = 7.24 × 10¹⁸ Hz
(b) Minimum wavelength = 4.14 × 10⁻¹¹ m or 0.0414 nm.
Question 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:
Work function:
φ₀ = 2.14 eV
Frequency:
ν = 6 × 10¹⁴ Hz
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
Mass of electron:
m = 9.1 × 10⁻³¹ kg
Charge of electron:
e = 1.6 × 10⁻¹⁹ C
Step 1. Calculate the energy of the incident photon
E = hν
E = 6.63 × 10⁻³⁴ × 6 × 10¹⁴
E = 3.978 × 10⁻¹⁹ J
Converting into electron volts:
E = (3.978 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
E = 2.49 eV
(a) Maximum kinetic energy
According to Einstein's photoelectric equation:
Kmax = hν − φ₀
Kmax = 2.49 − 2.14
Kmax = 0.35 eV
In joules:
Kmax = 0.35 × 1.6 × 10⁻¹⁹
Kmax = 5.6 × 10⁻²⁰ J
(b) Stopping potential
We know:
Kmax = eV₀
Therefore:
V₀ = Kmax/e
V₀ = 0.35 eV/1 e
V₀ = 0.35 V
(c) Maximum speed of photoelectrons
We know:
Kmax = ½mv²max
Therefore:
vmax = √(2Kmax/m)
Substituting:
vmax = √[(2 × 5.6 × 10⁻²⁰)/(9.1 × 10⁻³¹)]
vmax = 3.51 × 10⁵ m/s
Final Answer:
(a) Maximum kinetic energy = 0.35 eV or 5.6 × 10⁻²⁰ J
(b) Stopping potential = 0.35 V
(c) Maximum speed = 3.51 × 10⁵ m/s
Question 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:
V₀ = 1.5 V
Charge of electron:
e = 1.6 × 10⁻¹⁹ C
Formula:
Kmax = eV₀
Substituting:
Kmax = 1.6 × 10⁻¹⁹ × 1.5
Kmax = 2.4 × 10⁻¹⁹ J
Converting into electron volts:
Kmax = 1.5 eV
Final Answer:
Maximum kinetic energy = 1.5 eV or 2.4 × 10⁻¹⁹ J.
Question 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 the beam to have uniform cross-section which is less than the target area.
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Solution:
Given:
Wavelength:
λ = 632.8 nm
λ = 632.8 × 10⁻⁹ m
Power:
P = 9.42 mW
P = 9.42 × 10⁻³ W
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
Speed of light:
c = 3 × 10⁸ m/s
Mass of hydrogen atom:
mH = 1.67 × 10⁻²⁷ kg
(a) Energy and momentum of each photon
Energy of a photon:
E = hc/λ
E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(632.8 × 10⁻⁹)
E = 3.14 × 10⁻¹⁹ J
In electron volts:
E = (3.14 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)
E ≈ 1.96 eV
Momentum of a photon:
p = h/λ
p = (6.63 × 10⁻³⁴)/(632.8 × 10⁻⁹)
p = 1.05 × 10⁻²⁷ kg m/s
(b) Number of photons arriving per second
Power is the energy emitted per second.
If N is the number of photons emitted per second:
P = NE
Therefore:
N = P/E
N = (9.42 × 10⁻³)/(3.14 × 10⁻¹⁹)
N = 3 × 10¹⁶ photons/s
(c) Speed of hydrogen atom
The momentum of the hydrogen atom must be equal to the momentum of the photon.
Therefore:
mHv = p
v = p/mH
v = (1.05 × 10⁻²⁷)/(1.67 × 10⁻²⁷)
v = 0.63 m/s
Final Answer:
(a) Energy = 3.14 × 10⁻¹⁹ J; momentum = 1.05 × 10⁻²⁷ kg m/s
(b) Number of photons = 3 × 10¹⁶ photons/s
(c) Speed of hydrogen atom = 0.63 m/s
Question 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:
Given:
Slope of V₀ versus ν graph:
Slope = 4.12 × 10⁻¹⁵ V s
Charge of electron:
e = 1.6 × 10⁻¹⁹ C
Einstein's photoelectric equation is:
eV₀ = hν − φ₀
Dividing by e:
V₀ = (h/e)ν − φ₀/e
Comparing this equation with the equation of a straight line:
y = mx + c
Slope = h/e
Therefore:
h = e × Slope
Substituting:
h = 1.6 × 10⁻¹⁹ × 4.12 × 10⁻¹⁵
h = 6.592 × 10⁻³⁴ J s
Final Answer:
Planck's constant = 6.59 × 10⁻³⁴ J s.
Question 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
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
Charge of electron:
e = 1.6 × 10⁻¹⁹ C
Einstein's equation in terms of stopping potential is:
eV₀ = h(ν − ν₀)
Therefore:
V₀ = h(ν − ν₀)/e
Substituting:
V₀ = [6.63 × 10⁻³⁴ × (8.2 − 3.3) × 10¹⁴]/(1.6 × 10⁻¹⁹)
V₀ = [6.63 × 4.9 × 10⁻²⁰]/(1.6 × 10⁻¹⁹)
V₀ ≈ 2.03 V
Final Answer:
Stopping potential ≈ 2.0 V.
Question 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
We know:
Energy of photon:
E = hc/λ
Using the convenient relation:
E (in eV) = 1240/λ (in nm)
Substituting:
E = 1240/330
E = 3.76 eV
The energy of the incident photon is 3.76 eV.
Work function = 4.2 eV.
Since:
3.76 eV < 4.2 eV
The energy of the photon is less than the minimum energy required to remove an electron.
Therefore, photoelectric emission will not take place.
Final Answer:
No, the metal will not show photoelectric emission because the photon energy (3.76 eV) is less than its work function (4.2 eV).
Question 11.8
Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Solution:
Given:
Incident frequency:
ν = 7.21 × 10¹⁴ Hz
Maximum speed of photoelectrons:
vmax = 6.0 × 10⁵ m/s
Mass of electron:
m = 9.1 × 10⁻³¹ kg
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
According to Einstein's photoelectric equation:
hν = hν₀ + Kmax
Therefore:
hν₀ = hν − Kmax
We know:
Kmax = ½mv²max
Substituting:
Kmax = ½ × 9.1 × 10⁻³¹ × (6.0 × 10⁵)²
Kmax = 1.638 × 10⁻¹⁹ J
Energy of incident photon:
hν = 6.63 × 10⁻³⁴ × 7.21 × 10¹⁴
hν = 4.78 × 10⁻¹⁹ J
Therefore:
hν₀ = 4.78 × 10⁻¹⁹ − 1.638 × 10⁻¹⁹
hν₀ = 3.142 × 10⁻¹⁹ J
Threshold frequency:
ν₀ = (3.142 × 10⁻¹⁹)/(6.63 × 10⁻³⁴)
ν₀ ≈ 4.74 × 10¹⁴ Hz
Final Answer:
Threshold frequency ≈ 4.7 × 10¹⁴ Hz.
Question 11.9
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Solution:
Given:
Wavelength:
λ = 488 nm
Stopping potential:
V₀ = 0.38 V
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
Speed of light:
c = 3 × 10⁸ m/s
Energy of incident photon:
E = hc/λ
Using the relation:
E (in eV) = 1240/λ (in nm)
E = 1240/488
E = 2.54 eV
Maximum kinetic energy:
Kmax = eV₀
In electron volts:
Kmax = 0.38 eV
According to Einstein's photoelectric equation:
E = φ₀ + Kmax
Therefore:
φ₀ = E − Kmax
φ₀ = 2.54 − 0.38
φ₀ = 2.16 eV
Final Answer:
Work function = 2.16 eV.
Question 11.10
What is the de Broglie wavelength of:
(a) A bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.
(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s.
(c) A dust particle of mass 1.0 × 10⁻⁹ kg drifting with a speed of 2.2 m/s?
Solution:
According to de Broglie's relation:
λ = h/mv
Where:
h = 6.63 × 10⁻³⁴ J s
m = Mass of particle
v = Velocity of particle
(a) Bullet
Given:
m = 0.040 kg
v = 1.0 km/s = 1000 m/s
λ = h/mv
λ = (6.63 × 10⁻³⁴)/(0.040 × 1000)
λ = (6.63 × 10⁻³⁴)/40
λ = 1.66 × 10⁻³⁵ m
(b) Ball
Given:
m = 0.060 kg
v = 1.0 m/s
λ = h/mv
λ = (6.63 × 10⁻³⁴)/(0.060 × 1.0)
λ = 1.105 × 10⁻³² m
λ ≈ 1.1 × 10⁻³² m
(c) Dust particle
Given:
m = 1.0 × 10⁻⁹ kg
v = 2.2 m/s
λ = h/mv
λ = (6.63 × 10⁻³⁴)/(1.0 × 10⁻⁹ × 2.2)
λ = (6.63 × 10⁻³⁴)/(2.2 × 10⁻⁹)
λ ≈ 3.0 × 10⁻²⁵ m
Final Answer:
(a) Bullet: 1.66 × 10⁻³⁵ m
(b) Ball: 1.1 × 10⁻³² m
(c) Dust particle: 3.0 × 10⁻²⁵ m
These wavelengths are extremely small, which is why the wave nature of ordinary macroscopic objects is not noticeable.
Question 11.11
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
Let:
λ = Wavelength of electromagnetic radiation
ν = Frequency of electromagnetic radiation
h = Planck's constant
c = Speed of light
The energy of a photon is:
E = hν
The momentum of a photon is:
p = E/c
Substituting:
p = hν/c
We know:
c = νλ
Therefore:
p = hν/(νλ)
p = h/λ
According to de Broglie's relation:
λdB = h/p
Substituting p = h/λ:
λdB = h/(h/λ)
λdB = λ
Hence, the de Broglie wavelength of a photon is equal to the wavelength of the electromagnetic radiation.
3. IMPORTANT DERIVATIONS
Q1. Derive Einstein's photoelectric equation.
Answer:
According to Einstein, light consists of small energy packets called photons.
The energy of each photon is:
E = hν
When a photon falls on a metal surface, its energy is used in two ways:
A part of the energy is used to remove the electron from the metal surface. This minimum energy is called the work function (φ₀).
The remaining energy appears as the kinetic energy of the emitted electron.
According to the law of conservation of energy:
Energy of incident photon = Work function + Maximum kinetic energy
Therefore:
hν = φ₀ + Kmax
Hence:
Kmax = hν − φ₀
This is Einstein's photoelectric equation.
Since:
φ₀ = hν₀
We can also write:
Kmax = h(ν − ν₀)
Also:
Kmax = ½mv²max
Therefore:
½mv²max = hν − φ₀
In terms of stopping potential:
eV₀ = hν − φ₀
Q2. Derive the relation between stopping potential and frequency of incident light.
Answer:
Einstein's photoelectric equation is:
eV₀ = hν − φ₀
Dividing both sides by e:
V₀ = (h/e)ν − φ₀/e
This is in the form of a straight-line equation:
y = mx + c
Comparing:
y = V₀
x = ν
Slope = h/e
Intercept = −φ₀/e
Therefore:
V₀ = (h/e)ν − φ₀/e
The stopping potential increases linearly with the frequency of incident light above the threshold frequency.
Q3. Derive the de Broglie wavelength of a moving particle.
Answer:
According to de Broglie's hypothesis, every moving particle has an associated wavelength.
The wavelength is inversely proportional to the momentum of the particle.
Therefore:
λ ∝ 1/p
Introducing Planck's constant as the proportionality constant:
λ = h/p
For a particle of mass m moving with velocity v:
p = mv
Therefore:
λ = h/mv
This is the de Broglie wavelength equation.
Q4. Derive the de Broglie wavelength of an electron accelerated through a potential difference V.
Answer:
Let an electron of mass m and charge e be accelerated through a potential difference V.
The electrical energy gained by the electron is:
Energy = eV
This energy is converted into kinetic energy.
Therefore:
eV = ½mv²
Multiplying both sides by 2:
2eV = mv²
Multiplying by m:
2meV = m²v²
Taking square root:
mv = √(2meV)
But momentum:
p = mv
Therefore:
p = √(2meV)
According to de Broglie's equation:
λ = h/p
Substituting:
λ = h/√(2meV)
For an electron, this becomes:
λ = 1.227/√V nm
Where V is the accelerating potential in volts.
4. IMPORTANT QUESTIONS FOR EXAMINATION
Q1. Why is photoelectric emission an instantaneous process?
Answer: In Einstein's photon theory, one photon transfers its energy to one electron in a single interaction. If the photon energy is equal to or greater than the work function, the electron can be emitted immediately. Therefore, photoelectric emission occurs practically instantaneously.
Q2. Why does photoelectric current increase with the intensity of incident light?
Answer: Increasing the intensity of light increases the number of photons incident on the metal surface per second. More photons can release more electrons per second. Hence, photoelectric current increases.
Q3. Why does the maximum kinetic energy of photoelectrons depend on frequency and not on intensity?
Answer: The energy of each photon is E = hν. Increasing frequency increases the energy of each photon and hence the maximum kinetic energy of photoelectrons.
Increasing intensity at a fixed frequency increases the number of photons but does not increase the energy of each photon.
Q4. Why is there a threshold frequency for every metal?
Answer: Every metal has a definite work function, which is the minimum energy required to remove an electron from its surface.
Since photon energy is hν, the minimum frequency required is:
ν₀ = φ₀/h
This minimum frequency is called threshold frequency.
Q5. Can photoelectric emission take place if the intensity of light is very high but its frequency is below the threshold frequency?
Answer: No. If the frequency is below the threshold frequency, the energy of each photon is insufficient to overcome the work function. Increasing intensity increases the number of photons but does not increase the energy of an individual photon.
Therefore, photoelectric emission does not occur in the ordinary single-photon photoelectric effect.
Q6. What is the significance of the Davisson-Germer experiment?
Answer: The Davisson-Germer experiment demonstrated the diffraction of electrons by a nickel crystal. It provided experimental evidence for the wave nature of electrons and supported de Broglie's hypothesis.
Q7. Why is the de Broglie wavelength of a moving electron much more noticeable than that of a moving cricket ball?
Answer: An electron has a very small mass compared with a cricket ball. For comparable speeds, the electron has much smaller momentum and hence a larger de Broglie wavelength.
Therefore, its wave nature is easier to observe.
Q8. What happens to the de Broglie wavelength when the velocity of a particle is doubled?
Answer:
We know:
λ = h/mv
For constant mass:
λ ∝ 1/v
If velocity is doubled:
λ' = h/m(2v)
λ' = λ/2
Therefore, the de Broglie wavelength becomes half.
Q9. What happens to the de Broglie wavelength when the momentum of a particle is doubled?
Answer:
We know:
λ = h/p
Therefore:
λ ∝ 1/p
If momentum is doubled, the de Broglie wavelength becomes half.
Q10. What happens to the stopping potential when the frequency of incident light increases?
Answer: The stopping potential increases linearly with the frequency of incident light, provided the frequency is above the threshold frequency.
This follows from:
eV₀ = h(ν − ν₀)
Q11. What is the difference between photon and matter particle?
Answer:
| Photon | Matter Particle |
|---|---|
| It is a quantum of electromagnetic radiation. | It is a material particle such as an electron or proton. |
| Its rest mass is zero. | It has non-zero rest mass. |
| It travels at the speed of light in vacuum. | It travels at a speed less than the speed of light. |
| Its momentum is p = h/λ. | Its non-relativistic momentum is p = mv. |
| Its energy is E = hν. | Its kinetic energy at non-relativistic speeds is ½mv². |
5. IMPORTANT FORMULAS
Energy of photon:
E = hν
Energy of photon in terms of wavelength:
E = hc/λ
Momentum of photon:
p = h/λ
Momentum of photon in terms of energy:
p = E/c
Work function:
φ₀ = hν₀
Threshold wavelength:
λ₀ = c/ν₀
Einstein's photoelectric equation:
Kmax = hν − φ₀
Maximum kinetic energy:
Kmax = ½mv²max
Stopping potential relation:
Kmax = eV₀
Einstein's equation in terms of stopping potential:
eV₀ = h(ν − ν₀)
Stopping potential versus frequency:
V₀ = (h/e)ν − φ₀/e
Slope of V₀ versus ν graph:
Slope = h/e
Planck's constant from graph slope:
h = e × Slope
de Broglie wavelength:
λ = h/p
de Broglie wavelength for a non-relativistic particle:
λ = h/mv
Kinetic energy of a non-relativistic particle:
K = p²/2m
de Broglie wavelength in terms of kinetic energy:
λ = h/√(2mK)
de Broglie wavelength of an electron accelerated through potential V:
λ = h/√(2meV)
Approximate electron wavelength:
λ = 1.227/√V nm
Energy conversion:
1 eV = 1.6 × 10⁻¹⁹ J
6. IMPORTANT CONSTANTS
| Constant | Value |
|---|---|
| Planck's constant (h) | 6.63 × 10⁻³⁴ J s |
| Speed of light (c) | 3 × 10⁸ m/s |
| Charge of electron (e) | 1.6 × 10⁻¹⁹ C |
| Mass of electron (me) | 9.1 × 10⁻³¹ kg |
| Mass of proton (mp) | 1.67 × 10⁻²⁷ kg |
| 1 electron volt | 1.6 × 10⁻¹⁹ J |
7. IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Work function | Joule (J) |
| Photon energy | Joule (J) |
| Frequency | Hertz (Hz) |
| Wavelength | Metre (m) |
| Momentum | kg m/s |
| Kinetic energy | Joule (J) |
| Stopping potential | Volt (V) |
| Photoelectric current | Ampere (A) |
| Planck's constant | Joule-second (J s) |
| Velocity | Metre per second (m/s) |
| Refractive index | No unit |
8. QUICK REVISION POINTS
Light exhibits both wave and particle nature.
The particle nature of light is explained using photons.
The energy of a photon is E = hν.
The momentum of a photon is p = h/λ.
Work function is the minimum energy required to remove an electron from a metal surface.
Threshold frequency is the minimum frequency required for photoelectric emission.
Threshold wavelength is the maximum wavelength that can produce photoelectric emission.
Photoelectrons are electrons emitted due to incident radiation.
Stopping potential stops even the most energetic photoelectrons.
Saturation current is the maximum photoelectric current.
Increasing light intensity increases photoelectric current.
Increasing frequency increases the maximum kinetic energy of photoelectrons.
Below threshold frequency, ordinary single-photon photoelectric emission does not occur.
Einstein's photoelectric equation is Kmax = hν − φ₀.
The work function is related to threshold frequency by φ₀ = hν₀.
de Broglie proposed the wave nature of moving particles.
The de Broglie wavelength is λ = h/mv.
The Davisson-Germer experiment confirmed the wave nature of electrons.
The wave nature of macroscopic objects is difficult to observe because their de Broglie wavelengths are extremely small.
An electron accelerated through a potential difference V has wavelength λ = h/√(2meV).
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