RBSE Solutions For Class 12 Physics Chapter 13: Photoelectric Effect and Matter Waves | Textbook Important Questions & Answers

The answers are provided for all the questions of Chapter 13 Physics of RBSE Class 12. Students can go through these questions to understand the concepts better and score well in the board examination and entrance examinations for various professional courses. These solutions are provided by a team of experts and offer the best and accurate solutions to the questions. Do check BYJU’S RBSE Class 12 solutions page to get the answers for textbook questions of all subjects.

Multiple Choice Questions

Q1: A photon of energy 40eV is incident on a surface of the metal. Due to this, an electron is emitted of kinetic energy 37.5 eV. Find out the work function of the metal:

  1. 2.5 eV
  2. 57.5 eV
  3. 5.0 eV
  4. zero

Answer: (a)

Q2: In an experiment of photoelectric effect, the number of emitted photoelectrons for a light of frequency greater than threshold frequency is proportional to

  1. Its kinetic energy
  2. Its potential energy
  3. The frequency of the incident light
  4. The number of incident photons on metal

Answer:(d)

Q3: The energy of a photon of light beam A is twice the energy of a photon of light beam B. The ratio of their momentum PA/PB.

  1. ½
  2. ¼
  3. 4
  4. 2

Answer: ( c)

Q4: Electrons are emitted from a metal surface on the incidence of green colour of light. Among the following group of colours, which group will emit electrons?

  1. Yellow, Blue, Red
  2. Violet, Red, Yellow
  3. Violet, Blue, Yellow
  4. Violet, Blue, Sky Blue

Answer: (d)

Q5: ‘de- Broglie wavelength’ of an electron emitted from an electron gun is 0.1277 Å. The accelerating voltage of the gun is

  1. 20 kV
  2. 10 kV
  3. 30 kV
  4. 40 kV

Answer: (b)

Very Short Answers

Q1: Write Einstein’s photoelectric equation.

Answer: The equation is

hv = (½)mv2max + Φ

here hv = energy of incident light

Φ = work function

(½)mv2max= maximum kinetic energy of emitted photoelectrons

Q2: The stopping potential depends upon what?

Answer: The stopping potential depends only on the frequency of incident light.

Q3: To observe Photo-electric effect the frequency of incident light should be more than which frequency?

Answer: To observe Photo-electric effect the frequency of incident light should be greater than the threshold frequency of the material.

Q4: What is the name given to the quanta of electromagnetic energy?

Answer: The name given is Photon.

Q5: Write the formula for the wavelength of a matter-wave according to de-Broglie hypothesis

Answer: The wavelength is given by

λ = h/mv

h is the Planck’s constant

p = mv (momentum)

Q6: Write down the relationship between the uncertainties in the position of a particle and its associated momentum according to Heisenberg’s Uncertainty principle.

Answer: Δx. Δp ≥ h/2

Q7: Write the name of an experiment that establishes matter wave theory of de-Broglie

Answer: Davisson and Germer Experiment

Short Answer Type Questions

Q1: What is photoelectric effect?

Answer: When the light of a certain frequency illuminates a metal surface, electrons are emitted from the surface. This phenomenon is called the photoelectric effect.

Q2: What do you mean by Threshold frequency?

Answer: The minimum required frequency of the incident radiation to emit an electron from the surface is called threshold frequency.

Q3: Write down the definition of work function?

Answer: The minimum energy required by an electron to escape from the metal surface.

Q4: State the objective of Davisson Germer experiment.

Answer: Davisson Germer experiment is to study the wave properties of electrons.

Q5: Write down the hypothesis of de-Broglie about the dual nature of matter waves.

Answer: According to the hypothesis of de-Broglie, moving particles will have a wave associated with it like light and that is called matter waves. He said that the wavelength λ associated with a particle of momentum p is given by

λ = h/p, h is the Planck’s constant

Q6: Define the Uncertainty Principle

Answer: Heisenberg’s Uncertainty Principle states at any instant the position of a particle and its momentum in the same direction cannot be determined accurately at the same time and in the same direction.

ΔxΔpx ≥ ћ/2

here ћ = h/2π

= 1.054 x 1034 J.s

Δx is the uncertainty in the position of the particle

Δpx is the uncertainty in the momentum

Essay Type Questions

Q1: Explain the concept of photon and describe its various properties.

Answer:

The emission of free electrons from a metal surface when the light is incident on it is called the photoemission or the photoelectric effect. This effect led to the conclusion that light is made up of packets or quantum of energy called Photons

  • A photon is an elementary particle.
  • The momentum and energy of the photons are related as given below

E = p.c where

p = magnitude of the momentum

c = speed

  • Irrespective of the intensity of radiation, every photon of a frequency v has the same momentum.
  • The increase in the intensity of light only increases the number of photons crossing an area per unit time. It does not affect the energy of the radiation.
  • A photon remains unaffected by electric and magnetic fields. It is electrically neutral.
  • A photon is massless.
  • They are a stable particle.
  • The total energy and momentum are conserved during a photon-electron collision.
  • A photon cannot decay on its own.
  • The energy of a photon can be transferred during an interaction with other particles.
  • A photon has a spin-1, unlike electrons which are ½ spin. Its spin axis is parallel to the direction of travel. It is this property of photons which supports the polarization of light.

Q2: Explain the concept of the photoelectric effect by Einstein. What is meant by threshold frequency?

Answer: The light rays after striking the surface of the metal surface will collide with the free electrons of the metal. During certain collisions, the photon transfers the energy to the electron. If the energy is more than work function(Φ) then the electrons are emitted. It is not necessary that all the electrons will gain energy and get emitted from the surface. Some of the electrons in the metal may gain energy from photons and then collide with the ions and lose their energy. The maximum kinetic energy is given by E – Φ. The minimum energy may be zero.

Kmax = E – Φ

But energy of photon having frequency is given by

E = hν

Where h = Planck’s constant = 6.6261 × 10-34 Js.

Kmax = hv – Φ

hv =Φ + Kmax

This is the photoelectric equation of Einstein. This is also the statement of conservation of energy about work function Φ and absorption of a single photon. If the mass of the ejected electron is m and the maximum velocity is vmax we have. Thus, the maximum kinetic energy of the equation becomes:

Kmax = (½)mv2max

hv =Φ + (½)mv2max

If the stopping potential is v0 then Kmax = eV0

Hence hv =Φ + eV0

The explanation for experimental results of photoelectric effect using photoelectric equation

  1. The maximum kinetic energy Kmax varies linearly with the frequency of the incident light and not on the intensity.
  2. By definition, kinetic energy can never be negative. Hence, the photoelectric effect can be observed only when hv > Φ or hv > hv0

Here v0 = Φ/h

Therefore, the threshold frequency comes into existence. Light having a frequency less than the threshold cannot eject electrons whatever be its intensity.

  1. The intensity of light is proportional to the number of photons. If more photons are incident they will eject more photoelectrons.
  2. There is no time lag between incidence of light and emission of electrons.

All the results from the equation were in agreement with experimental observation.

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