Photoelectric Effect — Observations — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Photoelectric Effect — Observations 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 — Photoelectric Effect — Observations · easy · theory
Photoelectric emission from a metal surface is:
A. Delayed by several minutes at low intensity
B. Observed only after prolonged illumination
C. Delayed by several seconds at low intensity
D. Practically instantaneous, with no measurable time lag ✓ Correct
Solution: Emission begins within about $10^{-9}\text{ s}$, which the wave theory could never explain for feeble light.
Q2 — Photoelectric Effect — Observations · easy · theory
If the frequency of the incident light is below the threshold frequency, photoemission:
A. Occurs only for metals of high work function
B. Does not occur however intense the light is ✓ Correct
C. Occurs if the light is made intense enough
D. Occurs after a long time delay
Solution: A single photon must carry enough energy on its own; piling up more photons does not help.
Q3 — Photoelectric Effect — Observations · easy · theory
For light of frequency above threshold, the photoelectric current is:
A. Proportional to the square of the intensity
B. Directly proportional to the intensity of the incident light ✓ Correct
C. Independent of the intensity
D. Inversely proportional to the intensity
Solution: Greater intensity means more photons per second and therefore more ejected electrons per second.
Q4 — Photoelectric Effect — Observations · medium · theory
The maximum kinetic energy of photoelectrons depends on:
A. Neither frequency nor intensity
B. The frequency of the incident light, not its intensity ✓ Correct
C. Both the frequency and the intensity equally
D. The intensity of the incident light, not its frequency
Solution: Each electron absorbs a single photon, so its energy is fixed by that photon's frequency alone.
Q5 — Photoelectric Effect — Observations · medium · theory
The stopping potential for a given metal is:
A. Directly proportional to the intensity
B. Independent of the intensity of the incident light ✓ Correct
C. Inversely proportional to the intensity
D. Zero for all intensities
Solution: Stopping potential measures the maximum electron energy, which intensity does not affect.
Q6 — Photoelectric Effect — Observations · medium · theory
A graph of stopping potential against frequency of incident light is:
A. A horizontal straight line
B. A parabola
C. A straight line of slope $\dfrac{h}{e}$ ✓ Correct
D. A hyperbola
Solution: From $eV_0 = h\nu - \phi_0$, the intercept on the frequency axis gives the threshold frequency.
Q7 — Photoelectric Effect — Observations · medium · theory
The saturation photoelectric current depends on:
A. The intensity of the incident radiation ✓ Correct
B. The applied stopping potential
C. The frequency of the incident radiation
D. The work function of the metal only
Solution: At saturation every emitted electron reaches the collector, so the current simply counts the photons arriving.
Q8 — Photoelectric Effect — Observations · medium · theory
The classical wave theory of light fails to explain the photoelectric effect because it cannot account for:
A. The refraction of light in glass
B. The proportionality of current to intensity
C. The existence of a threshold frequency and the instantaneous emission ✓ Correct
D. The reflection of light from metals
Solution: Wave theory predicts that any frequency would work given enough time and intensity, which experiment flatly contradicts.
Q9 — Photoelectric Effect — Observations · medium · numerical
The slope of the graph of stopping potential against frequency in a photoelectric experiment equals:
A. $\dfrac{e}{h}$
B. $hc$
C. $\dfrac{h}{e}$ ✓ Correct
D. $h$
Solution: Rearranging $eV_0 = h\nu - \phi_0$ gives $V_0 = \left(\dfrac{h}{e}\right)\nu - \dfrac{\phi_0}{e}$.
Q10 — Photoelectric Effect — Observations · easy · numerical
The intensity of light falling on a photocell is doubled, the frequency being unchanged. The photoelectric current:
A. Halves
B. Doubles ✓ Correct
C. Becomes four times
D. Remains unchanged
Solution: Twice as many photons arrive each second, so twice as many electrons are ejected.
Q11 — Photoelectric Effect — Observations · medium · numerical
The intensity of light falling on a photocell is doubled at constant frequency. The stopping potential:
A. Remains unchanged ✓ Correct
B. Doubles
C. Becomes four times
D. Halves
Solution: Stopping potential depends only on the photon energy, which is set by frequency.
Q12 — Photoelectric Effect — Observations · medium · numerical
The stopping potential in a photoelectric experiment is $2\text{ V}$. The maximum kinetic energy of the photoelectrons is:
A. $3.2 \times 10^{-19}\text{ J}$ ✓ Correct
B. $2 \times 10^{-19}\text{ J}$
C. $3.2 \times 10^{-18}\text{ J}$
D. $1.6 \times 10^{-19}\text{ J}$
Solution: $K_{max} = eV_0 = 1.6 \times 10^{-19} \times 2 = 3.2 \times 10^{-19}\text{ J}$, that is $2\text{ eV}$.
Q13 — Photoelectric Effect — Observations · easy · numerical
The stopping potential is $1.5\text{ V}$. The maximum kinetic energy of the photoelectrons, in electron volt, is:
A. $1.5\text{ eV}$ ✓ Correct
B. $3.0\text{ eV}$
C. $0.75\text{ eV}$
D. $1.6\text{ eV}$
Solution: In electron volt the maximum kinetic energy is numerically equal to the stopping potential in volt.
Q14 — Photoelectric Effect — Observations · easy · numerical
The saturation current in a photocell is $4\text{ mA}$ for a certain intensity. If the intensity is doubled, the saturation current becomes:
A. $16\text{ mA}$
B. $2\text{ mA}$
C. $8\text{ mA}$ ✓ Correct
D. $4\text{ mA}$
Solution: Saturation current is proportional to intensity.
Q15 — Photoelectric Effect — Observations · medium · numerical
In a photoelectric experiment the photocurrent falls to zero when the collector is held at a potential of:
A. $+V_0$
B. Any negative potential
C. Zero
D. $-V_0$, the stopping potential ✓ Correct
Solution: A retarding potential equal to the stopping potential just prevents even the fastest electrons from arriving.
Q16 — Photoelectric Effect — Observations · medium · numerical
The threshold frequency of a metal is $5 \times 10^{14}\text{ Hz}$. Light of frequency $4 \times 10^{14}\text{ Hz}$ falls on it. The result is:
A. Photoemission after a long delay
B. Photoemission only at high intensity
C. No photoemission, whatever the intensity ✓ Correct
D. Photoemission with small kinetic energy
Solution: Below the threshold frequency a photon simply lacks the energy to free an electron.
Q17 — Photoelectric Effect — Observations · medium · numerical
The stopping potential is $3\text{ V}$. The maximum kinetic energy of the photoelectrons is:
A. $1.6 \times 10^{-19}\text{ J}$
B. $4.8 \times 10^{-18}\text{ J}$
C. $3 \times 10^{-19}\text{ J}$
D. $4.8 \times 10^{-19}\text{ J}$ ✓ Correct
Solution: $K_{max} = eV_0 = 1.6 \times 10^{-19} \times 3 = 4.8 \times 10^{-19}\text{ J}$.
Q18 — Photoelectric Effect — Observations · easy · numerical
Two beams of the same frequency have intensities in the ratio $1 : 3$. The ratio of the saturation currents they produce is:
A. $1 : 3$ ✓ Correct
B. $1 : 1$
C. $1 : 9$
D. $3 : 1$
Solution: Saturation current is directly proportional to intensity.
Q19 — Photoelectric Effect — Observations · hard · numerical
The slope of the stopping potential against frequency graph is approximately:
A. $4.14 \times 10^{-15}\text{ V}\cdot\text{s}$ ✓ Correct
B. $3 \times 10^8\text{ V}\cdot\text{s}$
C. $6.63 \times 10^{-34}\text{ V}\cdot\text{s}$
D. $1.6 \times 10^{-19}\text{ V}\cdot\text{s}$
Solution: $\dfrac{h}{e} = \dfrac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} \approx 4.14 \times 10^{-15}\text{ V}\cdot\text{s}$.
Q20 — Photoelectric Effect — Observations · hard · numerical
A metal of threshold frequency $4 \times 10^{14}\text{ Hz}$ is illuminated by light of $6 \times 10^{14}\text{ Hz}$. The maximum kinetic energy of the photoelectrons is:
A. $3.98 \times 10^{-19}\text{ J}$
B. $2.65 \times 10^{-19}\text{ J}$
C. $1.33 \times 10^{-19}\text{ J}$ ✓ Correct
D. $6.63 \times 10^{-19}\text{ J}$
Solution: $K_{max} = h(\nu - \nu_0) = 6.63 \times 10^{-34} \times 2 \times 10^{14} \approx 1.33 \times 10^{-19}\text{ J}$.
Q21 — Photoelectric Effect — Observations · easy · numerical
In a photoelectric experiment the frequency of light is increased while the intensity is kept constant. The maximum kinetic energy of the photoelectrons:
A. Remains unchanged
B. Increases ✓ Correct
C. Decreases
D. Becomes zero
Solution: $K_{max} = h\nu - \phi_0$ grows linearly with frequency.