Einstein's Photoelectric Equation — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Einstein's Photoelectric Equation MCQs with step-by-step solutions (21 questions). Part of Dual Nature of Radiation and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Einstein's Photoelectric Equation · easy · theory
Einstein's photoelectric equation is:
A. $h\nu = \dfrac{\phi_0}{K_{max}}$
B. $h\nu = \phi_0 + K_{max}$ ✓ Correct
C. $h\nu = \phi_0 K_{max}$
D. $h\nu = \phi_0 - K_{max}$
Solution: The photon energy is shared between freeing the electron and giving it kinetic energy.
Q2 — Einstein's Photoelectric Equation · easy · theory
The maximum kinetic energy of a photoelectron in terms of the threshold frequency is:
A. $h\nu\nu_0$
B. $h(\nu + \nu_0)$
C. $\dfrac{h\nu}{\nu_0}$
D. $h(\nu - \nu_0)$ ✓ Correct
Solution: Since $\phi_0 = h\nu_0$, the excess photon energy above the work function appears as kinetic energy.
Q3 — Einstein's Photoelectric Equation · medium · theory
The relation between stopping potential and frequency is:
A. $eV_0 = h\nu_0$
B. $eV_0 = h\nu$
C. $eV_0 = h(\nu - \nu_0)$ ✓ Correct
D. $eV_0 = h(\nu + \nu_0)$
Solution: The stopping potential just cancels the maximum kinetic energy of the emitted electrons.
Q4 — Einstein's Photoelectric Equation · medium · theory
Einstein's photoelectric equation is essentially a statement of:
A. The wave nature of light
B. Conservation of momentum
C. Conservation of energy for a single photon-electron interaction ✓ Correct
D. Conservation of charge
Solution: The photon delivers its whole energy to one electron, and that energy is fully accounted for.
Q5 — Einstein's Photoelectric Equation · medium · theory
In the photoelectric effect, a single photon:
A. Is shared among several electrons
B. Loses only part of its energy to the electron
C. Can eject many electrons at once
D. Interacts with and ejects at most one electron ✓ Correct
Solution: This one-to-one interaction is what makes the emission instantaneous and frequency-dependent.
Q6 — Einstein's Photoelectric Equation · medium · theory
The instantaneous nature of photoemission is explained by the photon picture because:
A. The metal surface is very thin
B. Electrons are very light
C. The whole photon energy is delivered to one electron in a single event ✓ Correct
D. Light waves carry large energy
Solution: There is no need to wait for energy to accumulate, as the wave picture would require.
Q7 — Einstein's Photoelectric Equation · easy · theory
The existence of a threshold frequency is explained by the fact that:
A. The metal reflects low-frequency light
B. A photon must carry at least the work function energy to free an electron ✓ Correct
C. Light of low frequency is absorbed by the air
D. Electrons move too slowly at low frequency
Solution: Below $\nu_0$ the single-photon energy $h\nu$ is simply less than $\phi_0$.
Q8 — Einstein's Photoelectric Equation · easy · theory
Einstein received the Nobel Prize in Physics in 1921 principally for his explanation of the:
A. Structure of the atom
B. Photoelectric effect ✓ Correct
C. Theory of relativity
D. Brownian motion
Solution: The photoelectric work established the quantum nature of light beyond doubt.
Q9 — Einstein's Photoelectric Equation · easy · theory
The photoelectric effect provides direct evidence for the:
A. Particle nature of electromagnetic radiation ✓ Correct
B. Wave nature of electrons
C. Wave nature of electromagnetic radiation
D. Existence of the nucleus
Solution: Interference and diffraction show the wave side; the photoelectric effect shows the particle side.
Q10 — Einstein's Photoelectric Equation · hard · numerical
The threshold frequency of a metal is $\nu_0$. Light of frequency $3\nu_0$ gives stopping potential $V_1$ and light of $2\nu_0$ gives $V_2$. The ratio $V_1 : V_2$ is:
A. $3 : 2$
B. $4 : 1$
C. $1 : 2$
D. $2 : 1$ ✓ Correct
Solution: $eV_1 = h(3\nu_0 - \nu_0) = 2h\nu_0$ and $eV_2 = h(2\nu_0 - \nu_0) = h\nu_0$, so the ratio is $2 : 1$.
Q11 — Einstein's Photoelectric Equation · hard · numerical
Light of frequency $2\nu_0$ gives maximum kinetic energy $K_1$ and light of $5\nu_0$ gives $K_2$. The ratio $K_1 : K_2$ is:
A. $1 : 5$
B. $1 : 4$ ✓ Correct
C. $2 : 5$
D. $1 : 2$
Solution: $K_1 = h\nu_0$ and $K_2 = 4h\nu_0$, so $K_1 : K_2 = 1 : 4$.
Q12 — Einstein's Photoelectric Equation · easy · numerical
A photon of energy $5\text{ eV}$ falls on a metal of work function $2\text{ eV}$. The maximum kinetic energy of the photoelectron is:
A. $2.5\text{ eV}$
B. $7\text{ eV}$
C. $3\text{ eV}$ ✓ Correct
D. $2\text{ eV}$
Solution: $K_{max} = h\nu - \phi_0 = 5 - 2 = 3\text{ eV}$.
Q13 — Einstein's Photoelectric Equation · medium · numerical
A photon of energy $6\text{ eV}$ falls on a metal of work function $2\text{ eV}$. The stopping potential is:
A. $6\text{ V}$
B. $8\text{ V}$
C. $3\text{ V}$
D. $4\text{ V}$ ✓ Correct
Solution: $eV_0 = 6 - 2 = 4\text{ eV}$, so $V_0 = 4\text{ V}$.
Q14 — Einstein's Photoelectric Equation · hard · numerical
A metal of threshold frequency $5 \times 10^{14}\text{ Hz}$ is illuminated by light of $8 \times 10^{14}\text{ Hz}$. The maximum kinetic energy is:
A. $5.30 \times 10^{-19}\text{ J}$
B. $3.31 \times 10^{-19}\text{ J}$
C. $8.62 \times 10^{-19}\text{ J}$
D. $1.99 \times 10^{-19}\text{ J}$ ✓ Correct
Solution: $K_{max} = 6.63 \times 10^{-34} \times 3 \times 10^{14} \approx 1.99 \times 10^{-19}\text{ J}$.
Q15 — Einstein's Photoelectric Equation · medium · numerical
Light of photon energy $4.5\text{ eV}$ ejects electrons of maximum kinetic energy $1.5\text{ eV}$. The work function of the metal is:
A. $6\text{ eV}$
B. $3\text{ eV}$ ✓ Correct
C. $4.5\text{ eV}$
D. $1.5\text{ eV}$
Solution: $\phi_0 = h\nu - K_{max} = 4.5 - 1.5 = 3\text{ eV}$.
Q16 — Einstein's Photoelectric Equation · hard · numerical
For a metal of work function $2.5\text{ eV}$ illuminated by light of wavelength $400\text{ nm}$, the maximum kinetic energy is approximately ($hc = 1240\text{ eV}\cdot\text{nm}$):
A. $5.6\text{ eV}$
B. $3.1\text{ eV}$
C. $1.6\text{ eV}$
D. $0.6\text{ eV}$ ✓ Correct
Solution: Photon energy $= \dfrac{1240}{400} = 3.1\text{ eV}$, so $K_{max} = 3.1 - 2.5 = 0.6\text{ eV}$.
Q17 — Einstein's Photoelectric Equation · hard · numerical
Light of wavelength $300\text{ nm}$ falls on a metal of work function $2\text{ eV}$. The maximum kinetic energy is approximately:
A. $4.13\text{ eV}$
B. $1.13\text{ eV}$
C. $6.13\text{ eV}$
D. $2.13\text{ eV}$ ✓ Correct
Solution: Photon energy $= \dfrac{1240}{300} \approx 4.13\text{ eV}$, so $K_{max} = 4.13 - 2 = 2.13\text{ eV}$.
Q18 — Einstein's Photoelectric Equation · medium · numerical
The stopping potential for a metal of work function $3\text{ eV}$ is $2\text{ V}$. The energy of the incident photon is:
A. $3\text{ eV}$
B. $1\text{ eV}$
C. $5\text{ eV}$ ✓ Correct
D. $6\text{ eV}$
Solution: $h\nu = \phi_0 + eV_0 = 3 + 2 = 5\text{ eV}$.
Q19 — Einstein's Photoelectric Equation · easy · numerical
At the threshold frequency, the maximum kinetic energy of the emitted photoelectrons is:
A. Maximum
B. Equal to the work function
C. Zero ✓ Correct
D. Equal to $h\nu_0$
Solution: The photon energy is just enough to free the electron with nothing left over.
Q20 — Einstein's Photoelectric Equation · hard · numerical
Light of frequency $4\nu_0$ falls on a metal of threshold frequency $\nu_0$. Compared with light of frequency $2\nu_0$, the maximum kinetic energy becomes:
A. Twice as large
B. Four times as large
C. Three times as large ✓ Correct
D. Half as large
Solution: $K$ goes from $h\nu_0$ to $3h\nu_0$, so it is tripled.
Q21 — Einstein's Photoelectric Equation · medium · numerical
The work function of a metal is $2\text{ eV}$. Light of energy $2\text{ eV}$ falls on it. The photoelectrons emitted have:
A. Kinetic energy of $2\text{ eV}$
B. Zero kinetic energy ✓ Correct
C. Kinetic energy of $1\text{ eV}$
D. Kinetic energy of $4\text{ eV}$
Solution: The incident energy exactly equals the work function, so emission is only just possible.