JEE Physics Modern Physics MCQs Set 01

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Chapter-wise Objective Questions: Modern Physics

Navigate directly to the 50 objective questions for Modern Physics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Incorrect statement is
(a) Rest mass of photon is zero
(b) Photon is a particle of zero momentum
(c) Photon travels in vacuum with speed of light
(d) Photon may feel gravitation
Show Answer & Explanation

Answer: (b) Photon is a particle of zero momentum

Explanation:
1. A photon has momentum \( p = h/\lambda \), which is not zero.
2. The other three statements are true.

Teacher's Note:
1. Momentum of a photon is E/c.

Question 2: Which of the following is true for photon (where symbols stand for their usual meaning)?
(a) \( E = hc/\lambda \)
(b) \( E = \tfrac12mv^2 \)
(c) \( p = E/2c \)
(d) \( E = \tfrac12mc^2 \)
Show Answer & Explanation

Answer: (a) \( E = hc/\lambda \)

Explanation:
1. Photon energy \( E = h\nu = hc/\lambda \).
2. Its momentum is \( p = E/c \), not E/2c.

Teacher's Note:
1. A photon's energy is not ½mv².

Question 3: Light of wavelength 5000 Å is incident on a photosensitive surface at a rate of 10⁻⁸ J/s. The number of photons falling per second on the surface will be
(a) \( 2.5\times10^{10} \)
(b) \( 2.5\times10^{11} \)
(c) \( 2.5\times10^{12} \)
(d) \( 2.5\times10^{13} \)
Show Answer & Explanation

Answer: (a) \( 2.5\times10^{10} \)

Explanation:
1. Energy of one photon \( = hc/\lambda = 1.989\times10^{-25}/5\times10^{-7} = 3.98\times10^{-19} \) J.
2. \( N = 10^{-8}/3.98\times10^{-19} = 2.5\times10^{10} \) per second.

Teacher's Note:
1. Divide the power by the energy of one photon.

Question 4: Photon of energy 6 eV is incident on a metal having work function 2.1 eV. The stopping potential is
(a) −1.9 V
(b) −8.1 V
(c) −3.9 V
(d) none of these
Show Answer & Explanation

Answer: (c) −3.9 V

Explanation:
1. \( K_{max} = 6 - 2.1 = 3.9 \) eV.
2. The stopping potential is \( -3.9 \) V (retarding).

Teacher's Note:
1. The stopping potential equals K_max in volts.

Question 5: The number of photo electrons emitted for light of a frequency ν (higher than the threshold frequency ν₀) is proportional to
(a) frequency of light (ν)
(b) ν − ν₀
(c) threshold frequency (ν₀)
(d) intensity of light
Show Answer & Explanation

Answer: (d) intensity of light

Explanation:
1. The number of photons per second is proportional to intensity.
2. Each photon ejects at most one electron, so the number of electrons is proportional to intensity.

Teacher's Note:
1. Frequency decides energy, intensity decides number.

Question 6: Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is
(a) \( 3\times10^{19} \)
(b) \( 9\times10^{17} \)
(c) \( 3\times10^{16} \)
(d) \( 9\times10^{15} \)
Show Answer & Explanation

Answer: (c) \( 3\times10^{16} \)

Explanation:
1. Photon energy \( = 1.989\times10^{-25}/667\times10^{-9} = 2.98\times10^{-19} \) J.
2. \( N = 9\times10^{-3}/2.98\times10^{-19}\approx3\times10^{16} \) per second.

Teacher's Note:
1. Number per second is power over energy per photon.

Question 7: Surface of sodium is illuminated by a light of 6000 Å wavelength. Work function of sodium is 1.6 eV. Then minimum K.E. of emitted electrons is
(a) 0 eV
(b) 1.53 eV
(c) 2.46 eV
(d) 4.14 eV
Show Answer & Explanation

Answer: (a) 0 eV

Explanation:
1. Photon energy \( = 12400/6000 = 2.07 \) eV, so the maximum K.E. is 0.47 eV.
2. Electrons coming from deeper in the metal lose energy on the way out, and the minimum K.E. of emitted electrons is zero.

Teacher's Note:
1. Maximum K.E. is hν − φ. Minimum is zero.

Question 8: When a point source of monochromatic light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the photoelectric cell, 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 9 times, so the saturation current becomes 18/9 = 2 mA.

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

Question 9: When ultraviolet rays are incident on a metal plate then photoelectric effect does not occur, it occurs by incidence of
(a) infrared rays
(b) X–rays
(c) radio wave
(d) light wave
Show Answer & Explanation

Answer: (b) X–rays

Explanation:
1. UV rays are below the threshold frequency of the metal.
2. X-rays have a higher frequency than UV, so they can eject electrons.

Teacher's Note:
1. Choose the radiation of higher frequency.

Question 10: Threshold wavelength for photoelectric emission from a metal surface is 5200 Å. Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from
(a) 1 W IR lamp
(b) 50 W UV lamp
(c) 50 W IR lamp
(d) 10 W IR lamp
Show Answer & Explanation

Answer: (b) 50 W UV lamp

Explanation:
1. Emission needs a wavelength below 5200 Å.
2. UV radiation has a shorter wavelength, while IR is longer.
3. So only the UV lamp works.

Teacher's Note:
1. Power does not matter below threshold frequency.

Question 11: When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 6V₀. When the same surface is illuminated with light of wavelength 2λ, the stopping potential is 2V₀. The threshold wavelength of this surface for photoelectric effect is
(a) 6λ
(b) 4λ/3
(c) 4λ
(d) 8λ
Show Answer & Explanation

Answer: (c) 4λ

Explanation:
1. \( hc/\lambda - \phi = 6eV_0 \) and \( hc/2\lambda - \phi = 2eV_0 \).
2. Subtracting, \( hc/2\lambda = 4eV_0 \), so \( hc/\lambda = 8eV_0 \) and \( \phi = 2eV_0 \).
3. \( \lambda_0 = hc/\phi = 4\lambda \).

Teacher's Note:
1. Solve the two Einstein equations together.

Question 12: If the light of wavelength λ is incident on metal surface, the ejected fastest electron has speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be
(a) smaller than \( \sqrt{4/3}\,v \)
(b) greater than \( \sqrt{4/3}\,v \)
(c) 2v
(d) zero
Show Answer & Explanation

Answer: (b) greater than \( \sqrt{4/3}\,v \)

Explanation:
1. \( K = hc/\lambda - \phi \) and \( K' = \tfrac43(hc/\lambda) - \phi = \tfrac43K + \tfrac\phi3 \).
2. So \( K' > \tfrac43K \), and \( v' > \sqrt{4/3}\,v \).

Teacher's Note:
1. The work function does not scale with the energy.

Question 13: A photon of wavelength 4400 Å is passing through vacuum. The effective mass and momentum of the photon are respectively
(a) \( 5\times10^{-36} \) kg, \( 1.5\times10^{-27} \) kg·m/s
(b) \( 5\times10^{-35} \) kg, \( 1.5\times10^{-26} \) kg·m/s
(c) zero, \( 1.5\times10^{-26} \) kg·m/s
(d) \( 5\times10^{-36} \) kg, \( 1.67\times10^{-43} \) kg·m/s
Show Answer & Explanation

Answer: (a) \( 5\times10^{-36} \) kg, \( 1.5\times10^{-27} \) kg·m/s

Explanation:
1. Momentum \( p = h/\lambda = 6.63\times10^{-34}/4.4\times10^{-7} = 1.5\times10^{-27} \) kg·m/s.
2. Effective mass \( = p/c = 5\times10^{-36} \) kg.

Teacher's Note:
1. Effective mass of a photon is hν/c².

Question 14: When photons of energy hν fall on an aluminium plate (of work function E₀), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be
(a) K + E₀
(b) 2K
(c) K
(d) K + hν
Show Answer & Explanation

Answer: (d) K + hν

Explanation:
1. \( K = h\nu - E_0 \).
2. \( K' = 2h\nu - E_0 = K + h\nu \).

Teacher's Note:
1. Substitute the new frequency directly.

Question 15: The threshold frequency for a certain photosensitive metal is ν₀. When it is illuminated by light of frequency ν = 2ν₀, the maximum velocity of photoelectrons is v₀. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency ν = 5ν₀?
(a) \( \sqrt2\,v_0 \)
(b) \( 2v_0 \)
(c) \( 2\sqrt2\,v_0 \)
(d) \( 4v_0 \)
Show Answer & Explanation

Answer: (b) \( 2v_0 \)

Explanation:
1. At 2ν₀: \( \tfrac12mv_0^2 = h(2\nu_0 - \nu_0) = h\nu_0 \).
2. At 5ν₀: \( K = h(5\nu_0 - \nu_0) = 4h\nu_0 \).
3. K is 4 times larger, so the speed doubles: \( 2v_0 \).

Teacher's Note:
1. Speed scales as the square root of K.

Question 16: The work functions of tungsten and sodium are 5.06 eV and 2.53 eV respectively. If the threshold wavelength for sodium is 5896 Å, then the threshold wavelength for tungsten will be
(a) 11792 Å
(b) 5896 Å
(c) 4312 Å
(d) 2948 Å
Show Answer & Explanation

Answer: (d) 2948 Å

Explanation:
1. \( \lambda_0\propto1/\phi \).
2. \( \lambda_W = 5896\times2.53/5.06 = 2948 \) Å.

Teacher's Note:
1. A larger work function gives a smaller threshold wavelength.

Question 17: The stopping potential necessary to reduce the photoelectric current to zero
(a) is directly proportional to wavelength of incident light
(b) uniformly increases with the wavelength of incident light
(c) is directly proportional to frequency of incident light
(d) linearly increases with the frequency of incident light
Show Answer & Explanation

Answer: (d) linearly increases with the frequency of incident light

Explanation:
1. \( eV_0 = h\nu - \phi \).
2. So \( V_0 \) increases linearly with ν (a straight line with a non-zero intercept), but is not directly proportional to it.

Teacher's Note:
1. Proportional means the graph passes through the origin.

Question 18: A photoelectric cell is illuminated by a point source of light 4 m away. When the source is shifted to 8 m then
(a) each emitted electron carries one quarter of the initial energy
(b) number of electrons emitted per second is half the initial number
(c) each emitted electron carries half the initial energy
(d) number of electrons emitted per second is a quarter of the initial number
Show Answer & Explanation

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

Explanation:
1. Intensity falls as \( 1/r^2 \), so it becomes one quarter.
2. The number of electrons per second falls to a quarter, while the energy per electron does not change.

Teacher's Note:
1. Distance affects the number, not the energy.

Question 19: 4.5 mW of 400 nm light is directed at a photoelectric cell. If 0.10% of the incident photons produce photoelectrons, the current in the cell is
(a) 1.45 µA
(b) 0.48 µA
(c) 0.42 mA
(d) 0.32 mA
Show Answer & Explanation

Answer: (a) 1.45 µA

Explanation:
1. Photons per second \( = (4.5\times10^{-3})(4\times10^{-7})/(1.989\times10^{-25}) = 9.05\times10^{15} \).
2. Electrons per second \( = 10^{-3}\times9.05\times10^{15} = 9.05\times10^{12} \).
3. \( I = 9.05\times10^{12}\times1.6\times10^{-19}\approx1.45 \) µA.

Teacher's Note:
1. Current is the electron rate times the charge.

Question 20: A graph is plotted between photo current (I) and applied voltage (V). The current is zero when the applied voltage is −4 V. Then the maximum kinetic energy of the emitted electrons is
(a) 2 eV
(b) 4 eV
(c) 0 eV
(d) 1 eV
Show Answer & Explanation

Answer: (b) 4 eV

Explanation:
1. The stopping potential is 4 V.
2. \( K_{max} = eV_0 = 4 \) eV.

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

Question 21: In a photoelectric experiment, electrons are ejected from metals X and Y by light of intensity I and frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If Y has a greater work function than X, which one of the following graphs best illustrates the expected results?
(a) Two parallel straight lines of V against f, the line for Y starting at a higher frequency than that for X
(b) Two straight lines of different slopes with the line for Y steeper
(c) Two parallel straight lines, the line for Y starting at a lower frequency than that for X
(d) Two straight lines of different slopes with the line for X steeper
Show Answer & Explanation

Answer: (a) Two parallel straight lines of V against f, the line for Y starting at a higher frequency than that for X

Explanation:
1. \( eV_0 = hf - \phi \), so both graphs have the same slope h/e.
2. The larger work function of Y moves its threshold to a higher frequency.

Teacher's Note:
1. The slope is a universal constant.

Question 22: The stopping potential for a photosensitive metal is V₁ at frequency ν₁ and V₂ at frequency ν₂ (ν₂ > ν₁). The Planck's constant and work function of the metal are
(a) \( \dfrac{V_2+V_1}{\nu_2+\nu_1}e,\ \dfrac{V_2\nu_1+V_1\nu_2}{\nu_2+\nu_1}e \)
(b) \( \dfrac{V_2+V_1}{\nu_2-\nu_1}e,\ \dfrac{V_2\nu_1+V_1\nu_2}{\nu_2-\nu_1}e \)
(c) \( \dfrac{V_2-V_1}{\nu_2+\nu_1}e,\ \dfrac{V_2\nu_1-V_1\nu_2}{\nu_2+\nu_1}e \)
(d) \( \dfrac{V_2-V_1}{\nu_2-\nu_1}e,\ \dfrac{V_2\nu_1-V_1\nu_2}{\nu_2-\nu_1}e \)
Show Answer & Explanation

Answer: (d) \( \dfrac{V_2-V_1}{\nu_2-\nu_1}e,\ \dfrac{V_2\nu_1-V_1\nu_2}{\nu_2-\nu_1}e \)

Explanation:
1. \( eV_2 - eV_1 = h(\nu_2 - \nu_1) \), so \( h = e(V_2-V_1)/(\nu_2-\nu_1) \).
2. \( \phi = h\nu_1 - eV_1 = e\dfrac{V_2\nu_1 - V_1\nu_2}{\nu_2-\nu_1} \).

Teacher's Note:
1. Use two Einstein equations and subtract.

Question 23: A graph is plotted between maximum kinetic energy of electron with frequency of incident photon in photo electric effect. The slope of the curve will be
(a) charge of electron
(b) work function of metal
(c) Planck's constant
(d) ratio of Planck constant and charge of electron
Show Answer & Explanation

Answer: (c) Planck's constant

Explanation:
1. \( K_{max} = h\nu - \phi \).
2. The slope is Planck's constant h.

Teacher's Note:
1. For a graph of V₀ against ν the slope would be h/e.

Question 24: Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speed of emitted electrons will be
(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 1 : 5
Show Answer & Explanation

Answer: (b) 1 : 2

Explanation:
1. \( K_1 = 1 - 0.5 = 0.5 \) eV and \( K_2 = 2.5 - 0.5 = 2 \) eV.
2. \( v\propto\sqrt K \), so \( v_1/v_2 = \sqrt{0.5/2} = 1/2 \).

Teacher's Note:
1. Speed varies as √K.

Question 25: An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Photon energy \( = 2.75 + 10 = 12.75 \) eV.
2. \( 13.6(1 - 1/n^2) = 12.75 \) gives \( 1/n^2 = 1/16 \), so n = 4.

Teacher's Note:
1. Use the Bohr energy difference.

Question 26: Maximum velocity of emitted photo electrons from a metal surface is \( 1.2\times10^6 \) m/s. If specific charge of electron is \( 1.8\times10^{11} \) C/kg then stopping potential will be
(a) 2 V
(b) 3 V
(c) 4 V
(d) 6 V
Show Answer & Explanation

Answer: (c) 4 V

Explanation:
1. \( eV_0 = \tfrac12mv^2 \), so \( V_0 = v^2/(2e/m) \).
2. \( V_0 = (1.44\times10^{12})/(3.6\times10^{11}) = 4 \) V.

Teacher's Note:
1. Use the specific charge e/m directly.

Question 27: In a Young's double slit arrangement with y = 1 mm (the position of the first dark fringe from the centre), d = 0.24 mm and D = 1.2 m, the light emitted from the source S is incident on a photosensitive surface of work function 2.2 eV. The stopping potential for electrons ejected from the centre of the screen will be
(a) 0.9 V
(b) 0.5 V
(c) 0.4 V
(d) 0.1 V
Show Answer & Explanation

Answer: (a) 0.9 V

Explanation:
1. First dark fringe: \( y = \lambda D/2d \), so \( \lambda = 2dy/D = 2(0.24\times10^{-3})(10^{-3})/1.2 = 4\times10^{-7} \) m = 4000 Å.
2. Photon energy \( = 12400/4000 = 3.1 \) eV.
3. \( V_0 = 3.1 - 2.2 = 0.9 \) V.

Teacher's Note:
1. Find the wavelength from the position of the dark fringe.

Question 28: If given particles are moving with same velocity, then maximum de-Broglie wavelength for
(a) proton
(b) α-particle
(c) neutron
(d) β-particle
Show Answer & Explanation

Answer: (d) β-particle

Explanation:
1. \( \lambda = h/mv \) is largest for the smallest mass.
2. A β-particle (electron) is the lightest.

Teacher's Note:
1. Wavelength varies inversely with mass.

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