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Dual Nature of Radiation and Matter — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Dual Nature of Radiation and Matter MCQs with step-by-step solutions covering Photoelectric Effect, Particle Nature of Light–The Photon, Wave Nature of Light. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Photoelectric Effect · easy · theory
Photoelectric emission occurs only when the incident light has more than a certain minimum
A. wavelength
B. intensity
C. frequency ✓ Correct
D. power
Solution: Photoelectric emission requires $\nu \geq \nu_0$ (threshold frequency). The incident light must have more than a certain minimum frequency, regardless of intensity.
Q2 — Photoelectric Effect · easy · numerical
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
A. 1.3 V
B. 0.5 V ✓ Correct
C. 2.3 V
D. 1.8 V
Solution: The stopping potential corresponds to the maximum kinetic energy:
$eV_0 = K_{max} = 0.5$ eV $\Rightarrow V_0 = 0.5$ V
Q3 — Photoelectric Effect · easy · theory
The number of photoelectrons emitted for light of a frequency $\nu$ (higher than the threshold frequency $\nu_0$) is proportional to
A. $\nu - \nu_0$
B. threshold frequency ($\nu_0$)
C. intensity of light ✓ Correct
D. frequency of light ($\nu$)
Solution: The number of photoelectrons emitted depends only on the intensity of the incident light (one photon ejects one electron). Frequency decides their maximum kinetic energy, not their number.
Q4 — Photoelectric Effect · easy · theory
A photocell employs photoelectric effect to convert
A. change in the frequency of light into a change in electric voltage
B. change in the intensity of illumination into a change in photoelectric current ✓ Correct
C. change in the intensity of illumination into a change in the work function of the photocathode
D. change in the frequency of light into a change in the electric current
Solution: For a fixed frequency, the photoelectric current increases linearly with the intensity of the incident light. A photocell therefore converts a change in the intensity of illumination into a change in photoelectric current.
Q5 — Photoelectric Effect · easy · theory
When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by incidence of
A. infrared rays
B. X-rays ✓ Correct
C. radiowaves
D. light waves
Solution: If ultraviolet light cannot cause emission, the incident wavelength must be even shorter (higher energy). Among the options, only X-rays have a wavelength shorter than ultraviolet, so only X-rays can cause photoelectric emission.
Q6 — Photoelectric Effect · easy · theory
Which of the following is not the property of cathode rays?
A. It produces heating effect
B. It does not deflect in electric field ✓ Correct
C. It casts shadow
D. It produces fluorescence
Solution: Cathode rays are streams of electrons (negatively charged), so they DO deflect in an electric field — towards the positive plate. They produce heating, travel in straight lines (cast shadows), and cause fluorescence.
Q7 — Photoelectric Effect · easy · numerical
A light source is at a distance d from a photoelectric cell, then the number of photoelectrons emitted from the cell is n. If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become
A. $\frac{n}{2}$
B. $2n$
C. $4n$ ✓ Correct
D. $n$
Solution: Intensity $\propto \frac{1}{d^2}$. Halving the distance makes the intensity 4 times, so the number of photoelectrons becomes $4n$.
Q8 — Photoelectric Effect · easy · theory
Einstein's work on photoelectric effect gives support to
A. $E = mc^2$
B. $E = h\nu$ ✓ Correct
C. $h\nu = \frac{1}{2}mv^2$
D. $E = \frac{h}{\lambda}$
Solution: Einstein explained the photoelectric effect by treating light as concentrated packets (photons), each of energy $h\nu$ — supporting the relation $E = h\nu$.
Q9 — Photoelectric Effect · easy · theory
When intensity of incident light increases
A. photocurrent increases ✓ Correct
B. photocurrent decreases
C. kinetic energy of emitted photoelectrons increases
D. kinetic energy of emitted photoelectrons decreases
Solution: More intensity means more incident photons, hence more ejected electrons and a larger photocurrent. The kinetic energy of the electrons depends on frequency, not intensity.
Q10 — Photoelectric Effect · easy · numerical
Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be
A. 0.58 eV ✓ Correct
B. 2.48 eV
C. 1.24 eV
D. 1.16 eV
Solution: $E = \frac{12375}{5000} = 2.48$ eV
$KE = E - W_0 = 2.48 - 1.9 = 0.58$ eV
Q11 — Photoelectric Effect · easy · theory
Which of the following is true?
A. The stopping potential increases with increasing intensity of incident light
B. The photocurrent increases with increasing intensity of light ✓ Correct
C. The current in photocell increases with increasing frequency of light
D. The photocurrent is proportional to applied voltage
Solution: Higher intensity means more photons per second, so more electrons are emitted and the photocurrent increases. Stopping potential depends on frequency (not intensity), and photocurrent does not depend on frequency or applied voltage.
Q12 — Photoelectric Effect · easy · numerical
If the threshold wavelength for a certain metal is 2000 Å, then the work function of the metal is
A. 6.2 J
B. 6.2 eV ✓ Correct
C. 6.2 MeV
D. 6.2 keV
Solution: $W_0 = \frac{hc}{\lambda_0} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}} = 9.9 \times 10^{-19}$ J $= 6.2$ eV
Q13 — Photoelectric Effect · easy · numerical
In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of −1.2 V. Then,
A. the energy of the incident photons is 4.7 eV ✓ Correct
B. the energy of the incident photons is 2.3 eV
C. if higher frequency photons be used, the photoelectric current will rise
D. when the energy of photons is 3.5 eV, the photoelectric current will be maximum
Solution: The stopping potential gives the maximum kinetic energy: $KE = 1.2$ eV
$h\nu = W_0 + KE = 3.5 + 1.2 = 4.7$ eV
Q14 — Photoelectric Effect · easy · theory
Number of ejected photoelectrons increases with increase
A. in intensity of light ✓ Correct
B. in wavelength of light
C. in frequency of light
D. Never
Solution: The number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation.
Q15 — Photoelectric Effect · easy · numerical
Ultraviolet radiation of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV)
A. $3.2 \times 10^{-21}$ J
B. $3.2 \times 10^{-19}$ J ✓ Correct
C. $7 \times 10^{-25}$ J
D. $9 \times 10^{-32}$ J
Solution: $KE = E - W_0 = 6.2 - 4.2 = 2.0$ eV $= 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$ J
Q16 — Photoelectric Effect · easy · numerical
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of 5000 Å. Its work function is
A. $4 \times 10^{-19}$ J ✓ Correct
B. 1 J
C. $2 \times 10^{-19}$ J
D. $3 \times 10^{-19}$ J
Solution: $W_0 = \frac{hc}{\lambda_0} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = 4 \times 10^{-19}$ J
Q17 — Photoelectric Effect · easy · theory
Which of the following are thermions?
A. Protons
B. Electrons ✓ Correct
C. Photons
D. Positrons
Solution: Thermionic emission is the emission of electrons from a metal surface when it is suitably heated. The emitted electrons are called thermions.
Q18 — Particle Nature of Light–The Photon · easy · theory
The nature of ions knocked out from hot surfaces is
A. protons
B. electrons ✓ Correct
C. neutrons
D. nuclei
Solution: Heating a metal surface causes thermionic emission — the particles knocked out are electrons.
Q19 — Particle Nature of Light–The Photon · easy · theory
Momentum of a photon of wavelength $\lambda$ is
A. $\frac{h}{\lambda}$ ✓ Correct
B. zero
C. $\frac{h\lambda}{c^2}$
D. $\frac{h\lambda}{c}$
Solution: A photon moves with speed c, so its momentum is
$p = \frac{h\nu}{c} = \frac{h}{\lambda}$
Q20 — Particle Nature of Light–The Photon · easy · theory
The energy of a photon of wavelength $\lambda$ is
A. $hc\lambda$
B. $\frac{hc}{\lambda}$ ✓ Correct
C. $\frac{\lambda}{hc}$
D. $\frac{\lambda h}{c}$
Solution: By Planck's quantum theory, the energy of a photon is
$E = h\nu = \frac{hc}{\lambda}$
Q21 — Wave Nature of Light · easy · theory
The wave nature of electrons was experimentally verified by
A. de-Broglie
B. Hertz
C. Einstein
D. Davisson and Germer ✓ Correct
Solution: de-Broglie proposed the hypothesis of matter waves, but it was the Davisson–Germer experiment (electron diffraction from a nickel crystal) that experimentally verified the wave nature of electrons.
Q22 — Wave Nature of Light · easy · theory
The following particles are moving with the same velocity, then maximum de-Broglie wavelength will be for
A. proton
B. $\alpha$-particle
C. neutron
D. $\beta$-particle ✓ Correct
Solution: $\lambda = \frac{h}{mv}$; for the same velocity, $\lambda \propto \frac{1}{m}$
The $\beta$-particle (an electron) has the smallest mass, so its de-Broglie wavelength is maximum.
Q23 — Wave Nature of Light · easy · numerical
The energy of a photon of light is 3 eV. Then the wavelength of photon must be
A. 4125 nm
B. 412.5 nm ✓ Correct
C. 41250 nm
D. 4 nm
Solution: $\lambda = \frac{12375}{E(\text{eV})}$ Å $= \frac{12375}{3} = 4125$ Å $= 412.5$ nm
Q24 — Wave Nature of Light · easy · theory
The de-Broglie wave corresponding to a particle of mass m and velocity v has a wavelength associated with it
A. $\frac{h}{mv}$ ✓ Correct
B. $hmv$
C. $\frac{mh}{v}$
D. $\frac{m}{hv}$
Solution: The de-Broglie wavelength of a moving particle is
$\lambda = \frac{h}{mv}$
where h is Planck's constant.
Q25 — Photoelectric Effect · hard · numerical
The photoelectric threshold wavelength of silver is $3250 \times 10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10}$ m is (Given, $h = 4.14 \times 10^{-15}$ eV-s and $c = 3 \times 10^8$ ms⁻¹)
A. $\approx 6 \times 10^5$ ms⁻¹ ✓ Correct
B. $\approx 0.6 \times 10^6$ ms⁻¹
C. $\approx 61 \times 10^3$ ms⁻¹
D. $\approx 0.3 \times 10^6$ ms⁻¹
Solution: $\frac{1}{2}m_e v^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$
$v = \sqrt{\frac{2hc}{m_e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)} \approx 6 \times 10^5$ ms⁻¹
(Note: options (a) and (b) are numerically the same value — $0.6 \times 10^6 = 6 \times 10^5$; the official key accepted both.)
Q26 — Photoelectric Effect · hard · numerical
When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is
A. $5\lambda$
B. $\frac{5}{2}\lambda$
C. $3\lambda$ ✓ Correct
D. $4\lambda$
Solution: Case 1: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
Case 2: $\frac{eV}{4} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$, i.e. $eV = \frac{4hc}{2\lambda} - \frac{4hc}{\lambda_0}$
Equating: $\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{4}{\lambda_0} \Rightarrow \lambda_0 = 3\lambda$
Q27 — Photoelectric Effect · hard · numerical
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
A. +3 V
B. +4 V
C. −1 V
D. −3 V ✓ Correct
Solution: Work function: $\phi = 5 - 2 = 3$ eV
With 6 eV photons, $K_{max} = 6 - 3 = 3$ eV.
To stop these electrons the anode must be 3 V negative relative to the cathode: $V_A - V_C = -3$ V
Q28 — Photoelectric Effect · hard · numerical
A certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface for photoelectric effect is
A. $6\lambda$
B. $4\lambda$ ✓ Correct
C. $\frac{\lambda}{4}$
D. $\frac{\lambda}{6}$
Solution: $\frac{hc}{\lambda} = W + 3eV_0$ and $\frac{hc}{2\lambda} = W + eV_0$ (so $\frac{hc}{\lambda} = 2W + 2eV_0$)
Subtracting: $W = eV_0$, hence $\frac{hc}{\lambda} = 4eV_0$
$\lambda_{th} = \frac{hc}{W} = \frac{4eV_0\lambda}{eV_0} = 4\lambda$
Q29 — Photoelectric Effect · hard · numerical
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is
A. $< 2.8 \times 10^{-10}$ m
B. $< 2.8 \times 10^{-9}$ m
C. $\geq 2.8 \times 10^{-9}$ m ✓ Correct
D. $\leq 2.8 \times 10^{-12}$ m
Solution: Photon energy $E = \frac{hc}{\lambda} = 2.48$ eV, so $K_{max} = 2.48 - 2.28 = 0.2$ eV
Minimum de-Broglie wavelength: $\lambda_e = \frac{12.27}{\sqrt{0.2}}$ Å $\approx 2.8 \times 10^{-9}$ m
Slower electrons have longer wavelengths, so $\lambda \geq 2.8 \times 10^{-9}$ m.
Q30 — Photoelectric Effect · hard · numerical
In a photoemissive cell, with exciting wavelength $\lambda$, the fastest electron has speed v. If the exciting wavelength is changed to $\frac{3\lambda}{4}$, the speed of the fastest emitted electron will be
A. $v\left(\frac{3}{4}\right)^{1/2}$
B. $v\left(\frac{4}{3}\right)^{1/2}$
C. less than $v\left(\frac{4}{3}\right)^{1/2}$
D. greater than $v\left(\frac{4}{3}\right)^{1/2}$ ✓ Correct
Solution: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - W_0$ and $\frac{1}{2}mv'^2 = \frac{4hc}{3\lambda} - W_0$
Since the photon energy increases by a factor $\frac{4}{3}$ but the work function stays the same, the kinetic energy increases by more than $\frac{4}{3}$:
$v' > v\sqrt{\frac{4}{3}}$