JEE Physics Dual Nature Of Radiation And Matter MCQs Set 04

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Multiple Choice Questions

Question 1: The energy of a photon of visible light of maximum wavelength is about (in eV)
(a) 1
(b) 1.6
(c) 3.2
(d) 7
Show Answer & Explanation

Answer: (b) 1.6

Explanation:
1. Visible light extends to about 7000 Å, so \( E=1240/700\approx1.8 \) eV.
2. The nearest option is 1.6 eV.

Teacher's Note:
1. Red end of the visible spectrum.

Question 2: The momentum of a photon of frequency \( 1.5\times10^{13} \) Hz is
(a) \( 3.3\times10^{-29}\ \text{kg m/s} \)
(b) \( 3.3\times10^{-34}\ \text{kg m/s} \)
(c) \( 6.6\times10^{-34}\ \text{kg m/s} \)
(d) \( 6.6\times10^{-30}\ \text{kg m/s} \)
Show Answer & Explanation

Answer: (a) \( 3.3\times10^{-29}\ \text{kg m/s} \)

Explanation:
1. \( p=h\nu/c=6.63\times10^{-34}\times1.5\times10^{13}/(3\times10^{8}) \).
2. \( p=3.3\times10^{-29} \) kg m/s.

Teacher's Note:
1. p = h nu / c.

Question 3: The strength of the photoelectric current is directly proportional to
(a) Frequency of incident radiation
(b) Intensity of incident radiation
(c) Angle of incidence of radiation
(d) Distance between anode and cathode
Show Answer & Explanation

Answer: (b) Intensity of incident radiation

Explanation:
1. Photocurrent depends on the number of photons per second, which is proportional to the intensity.

Teacher's Note:
1. Saturation current is proportional to intensity.

Question 4: When light is incident on a surface, photoelectrons are emitted. For these photoelectrons
(a) The value of kinetic energy is the same for all
(b) Maximum kinetic energy does not depend on the wavelength of incident light
(c) The kinetic energy is equal to or less than a maximum kinetic energy
(d) None of the above
Show Answer & Explanation

Answer: (c) The kinetic energy is equal to or less than a maximum kinetic energy

Explanation:
1. Energy lost inside the metal gives a spread of energies from zero up to a maximum.

Teacher's Note:
1. Maximum KE = h nu - phi.

Question 5: The kinetic energy of photoelectrons does not depend on
(a) Wavelength of light
(b) Frequency of light
(c) Type of material used for the surface
(d) Intensity of light
Show Answer & Explanation

Answer: (d) Intensity of light

Explanation:
1. Kinetic energy depends on the frequency and the work function, not on the intensity.

Teacher's Note:
1. Intensity affects the number only.

Question 6: The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission is about
(a) 540 nm
(b) 400 nm
(c) 310 nm
(d) 220 nm
Show Answer & Explanation

Answer: (c) 310 nm

Explanation:
1. \( \lambda_{max}=1240/4.0=310 \) nm.

Teacher's Note:
1. Threshold wavelength = hc / phi.

Question 7: Photoelectric effect takes place in element A with work function 2.5 eV and threshold wavelength \( \lambda \). Element B has work function 5 eV. The maximum wavelength that can produce photoelectric effect in B is
(a) \( \lambda/2 \)
(b) \( 2\lambda \)
(c) \( \lambda \)
(d) \( 3\lambda \)
Show Answer & Explanation

Answer: (a) \( \lambda/2 \)

Explanation:
1. \( \lambda_0=hc/\phi \), so doubling \( \phi \) halves the wavelength.

Teacher's Note:
1. Inverse proportion.

Question 8: When light of wavelength less than 6000 Å is incident on a metal, electrons are emitted. The approximate work function of the metal is
(a) 1 eV
(b) 2 eV
(c) 4 eV
(d) 6 eV
Show Answer & Explanation

Answer: (b) 2 eV

Explanation:
1. \( \phi=1240/600\approx2.07 \) eV.

Teacher's Note:
1. Threshold wavelength is 6000 Å.

Question 9: The maximum kinetic energy of photoelectrons from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is
(a) 4 V
(b) 6 V
(c) 8 V
(d) 10 V
Show Answer & Explanation

Answer: (a) 4 V

Explanation:
1. \( eV_0=K_{max} \), so \( V_0=4 \) V.

Teacher's Note:
1. Stopping potential equals K max in volts.

Question 10: A point source of monochromatic light is 0.2 m from a photoelectric cell; the cut-off voltage is 0.6 V and the saturation current is 18 mA. If the source is placed 0.6 m away, then
(a) The stopping potential will be 0.2 V
(b) The stopping potential will be 0.6 V
(c) The saturation current will be 6 mA
(d) The saturation current will be 18 mA
Show Answer & Explanation

Answer: (b) The stopping potential will be 0.6 V

Explanation:
1. The stopping potential depends only on frequency, so it stays 0.6 V.
2. The intensity falls to 1/9, so the saturation current falls to 2 mA.

Teacher's Note:
1. Distance changes intensity not frequency.

Question 11: The maximum wavelength of light for photoelectric effect from a metal is 200 nm. The maximum kinetic energy of the electron emitted by radiation of wavelength 100 nm is
(a) 12.4 eV
(b) 6.2 eV
(c) 100 eV
(d) 200 eV
Show Answer & Explanation

Answer: (b) 6.2 eV

Explanation:
1. \( \phi=1240/200=6.2 \) eV and \( E=1240/100=12.4 \) eV.
2. \( K=12.4-6.2=6.2 \) eV.

Teacher's Note:
1. K max = h nu - phi.

Question 12: The slope of the graph between the stopping potential and the frequency of incident light for a given surface is
(a) h
(b) h/e
(c) eh
(d) e
Show Answer & Explanation

Answer: (b) h/e

Explanation:
1. \( eV_0=h\nu-\phi \), so \( V_0=(h/e)\nu-\phi/e \).

Teacher's Note:
1. Slope h/e.

Question 13: By the photoelectric effect, Einstein proved
(a) \( E=h\nu \)
(b) \( KE=\tfrac12mv^2 \)
(c) \( E=mc^2 \)
(d) \( E=-Rhc^2/n^2 \)
Show Answer & Explanation

Answer: (a) \( E=h\nu \)

Explanation:
1. Light energy is in quanta of energy hν.

Teacher's Note:
1. Quantum nature of light.

Question 14: Which of the following shows the particle nature of light?
(a) Photoelectric effect
(b) Interference
(c) Refraction
(d) Polarisation
Show Answer & Explanation

Answer: (a) Photoelectric effect

Explanation:
1. The photoelectric effect can only be explained by photons.

Teacher's Note:
1. Interference and polarisation show wave nature.

Question 15: A photocell is illuminated by a source at a distance d. If the distance becomes d/2, the number of electrons emitted per second will be
(a) Unchanged
(b) Four times
(c) Two times
(d) One-fourth
Show Answer & Explanation

Answer: (b) Four times

Explanation:
1. Intensity is proportional to \( 1/d^2 \), so it becomes four times.

Teacher's Note:
1. Photocurrent is proportional to intensity.

Question 16: A graph is plotted between the maximum kinetic energy of the electron and the frequency of the incident photon. The slope of the curve is
(a) Charge of electron
(b) Work function of metal
(c) Planck's constant
(d) Ratio of Planck constant to charge of electron
Show Answer & Explanation

Answer: (c) Planck's constant

Explanation:
1. \( K=h\nu-\phi \) has slope h.

Teacher's Note:
1. Slope h for K against frequency.

Question 17: A photon of energy 6 eV is incident on a metal of work function 4 eV. The maximum KE of the emitted photoelectrons is
(a) 0 eV
(b) 1 eV
(c) 2 eV
(d) 10 eV
Show Answer & Explanation

Answer: (c) 2 eV

Explanation:
1. \( K=6-4=2 \) eV.

Teacher's Note:
1. Einstein's equation.

Question 18: Light of frequency \( \nu \) is incident on a metal of threshold frequency \( \nu_0 \). The work function of the metal is
(a) \( h\nu \)
(b) \( h\nu_0 \)
(c) \( h(\nu-\nu_0) \)
(d) \( h(\nu+\nu_0) \)
Show Answer & Explanation

Answer: (b) \( h\nu_0 \)

Explanation:
1. \( \phi=h\nu_0 \).

Teacher's Note:
1. Work function is h times threshold frequency.

Question 19: The work function of a photoelectric material is 3.3 eV. Its threshold frequency is
(a) \( 4\times10^{23}\ \text{Hz} \)
(b) \( 8\times10^{12}\ \text{Hz} \)
(c) \( 4\times10^{11}\ \text{Hz} \)
(d) \( 8\times10^{14}\ \text{Hz} \)
Show Answer & Explanation

Answer: (d) \( 8\times10^{14}\ \text{Hz} \)

Explanation:
1. \( \nu_0=\phi/h=3.3\times1.6\times10^{-19}/6.63\times10^{-34} \).
2. \( \nu_0\approx8\times10^{14} \) Hz.

Teacher's Note:
1. nu0 = phi / h.

Question 20: The unit of Planck's constant is
(a) \( \text{J s} \)
(b) \( \text{J m}^{2}/\text{s} \)
(c) \( \text{kg m}^{-2} \)
(d) \( \text{kg s}^{-1} \)
Show Answer & Explanation

Answer: (a) \( \text{J s} \)

Explanation:
1. \( h=E/\nu \), so its unit is J s.

Teacher's Note:
1. Dimensions of angular momentum.

Question 21: Ultraviolet radiation falling on a metal does not cause photoelectric effect, but it occurs with
(a) Infrared rays
(b) X-rays
(c) Radio waves
(d) Visible light
Show Answer & Explanation

Answer: (b) X-rays

Explanation:
1. Only radiation of higher frequency than UV (X-rays) can have enough energy.

Teacher's Note:
1. Higher frequency means more energy.

Question 22: A photoelectric cell is illuminated by a point source 1 m away. When the source is shifted to 2 m
(a) Each emitted electron carries one quarter of the initial energy
(b) The number of electrons emitted is half the initial number
(c) Each emitted electron carries half the initial energy
(d) The number of electrons emitted is a quarter of the initial number
Show Answer & Explanation

Answer: (d) The number of electrons emitted is a quarter of the initial number

Explanation:
1. Intensity falls to 1/4, so the number of photoelectrons per second falls to 1/4.
2. Their energy is unchanged.

Teacher's Note:
1. Distance changes intensity only.

Question 23: According to Einstein's photoelectric equation, the slope of the line plotting the kinetic energy of emitted photoelectrons against the frequency of the incident radiation
(a) Depends on the intensity of the radiation
(b) Depends on the nature of the metal used
(c) Depends on both intensity and metal
(d) Is the same for all metals and independent of intensity
Show Answer & Explanation

Answer: (d) Is the same for all metals and independent of intensity

Explanation:
1. The slope equals Planck's constant.

Teacher's Note:
1. Slope is universal.

Question 24: A photon of energy 4 eV is incident on a metal of work function 2 eV. The minimum reverse potential for stopping the emission is
(a) 2 V
(b) 4 V
(c) 6 V
(d) 8 V
Show Answer & Explanation

Answer: (a) 2 V

Explanation:
1. \( K=4-2=2 \) eV, so the stopping potential is 2 V.

Teacher's Note:
1. Stopping potential in volts equals K in eV.

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