Q34

1 markMCQSection D

For radiation of frequency $\nu > \nu_2$ incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

Q: For radiation of frequency $\nu > \nu_2$ incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

**Correct option: (A) greater for metal A because it has a smaller work function.** Maximum kinetic energy: $K_{max} = h\nu - \phi$. - Metal A has the smaller threshold frequency $(\nu_1 < \nu_2)$, so $\phi_A = h\nu_1 < \phi_B = h\nu_2$. - For the **same** incident frequency ν, a smaller φ leaves more energy as kinetic energy. Hence $K_{max}$ is **greater for metal A**.

Q: Why does metal A give photoelectrons with more kinetic energy than metal B?

Maximum kinetic energy is $K_{max} = h\nu - \phi$. Metal A has the smaller threshold frequency $(\nu_1 < \nu_2)$, so its work function $\phi_A = h\nu_1$ is smaller. At the same incident frequency the two metals absorb the same photon energy $h\nu$, but A spends less of it escaping the surface, leaving more as kinetic energy. So $K_{max}$ is larger for A by exactly $\phi_B - \phi_A = h(\nu_2 - \nu_1)$.

Q: Does the maximum kinetic energy depend on the intensity of the light?

No. $K_{max}$ depends only on the frequency of the incident light and the metal's work function. Increasing intensity increases the number of photoelectrons per second but not their maximum kinetic energy — one of the key experimental facts explained only by the photon picture.

Graph of stopping potential vs frequency for metals A and B

Dual Nature of Radiation and Matter

Comparing maximum kinetic energy of photoelectrons

Options

(A)greater for metal A because it has a smaller work function.

(B)greater for metal B because it has a larger work function.

(C)greater for metal B because it has higher threshold frequency.

(D)the same for both metal A and metal B because it is independent of work functions of metals.

Official Answer

Correct option: (A) greater for metal A because it has a smaller work function.

Maximum kinetic energy: $K_{max} = h\nu - \phi$ .

Metal A has the smaller threshold frequency $(\nu_1 < \nu_2)$ , so $\phi_A = h\nu_1 < \phi_B = h\nu_2$ .
For the same incident frequency ν, a smaller φ leaves more energy as kinetic energy.

Hence $K_{max}$ is greater for metal A.

maximum kinetic energyK_max = hν − φsmaller work function larger KEthreshold frequency comparisonmetal A metal Bsame incident frequencyphotoelectric effect

Marking Scheme

11 mark: correct option (A) — greater for A because it has the smaller work function.
2Reasoning credit for $K_{max} = h\nu - \phi$ and identifying $\phi_A < \phi_B$ from $\nu_1 < \nu_2$ .
3No marks for options B, C (B greater) or D (independent of φ).

Hint

$K_{max} = h\nu - \phi$ at fixed ν; smaller work function ⇒ larger $K_{max}$ , and A has the smaller threshold frequency.

Quick Oral Answer

Since $K_{max}$ equals $h\nu$ minus the work function, at the same frequency the metal with the smaller work function keeps more kinetic energy; metal A has the lower threshold $\nu_1$ , so its photoelectrons are faster.

Analysis & Explanation

Concept:

Einstein's equation $K_{max} = h\nu - \phi$ shows that for a fixed incident frequency the photoelectron's maximum kinetic energy depends only on the work function: the smaller the work function, the larger the leftover kinetic energy.

Reading the graph:

Metal A intercepts the frequency axis at the smaller value $\nu_1$ , so its threshold frequency and work function are smaller.
At any common $\nu > \nu_2$ (so both metals emit), $K_{max}(A) - K_{max}(B) = \phi_B - \phi_A = h(\nu_2 - \nu_1) > 0$ .

Why the distractors fail:

(B), (C) claim B is greater — a larger work function or higher threshold means less kinetic energy, not more.
(D) is wrong because K_max clearly depends on φ; only the slope $(h/e)$ , not the intercept, is metal-independent.

Exam trap:

Do not confuse 'more energy needed to escape' with 'more energy left over' — a higher work function takes away kinetic energy.

Common Mistakes

1Thinking a larger work function gives a larger kinetic energy (it does the opposite).
2Assuming kinetic energy is independent of the metal because the graph lines are parallel — the intercept still shifts $K_{max}$ .
3Confusing threshold frequency with the energy of the emitted electron.

Interesting Facts

Because the two lines are parallel, the vertical gap between them — the difference in stopping potential at any frequency — is constant and equals $(\phi_B - \phi_A)/e$ .

Metals with low work functions like sodium (2.28 eV) and potassium (2.30 eV) release faster photoelectrons than high-work-function metals like platinum (about 6 eV) under the same light.

$K_{max}$ depends only on frequency and work function, never on intensity — a fact that sealed the case for the photon model over classical wave theory.

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Frequently Asked Questions

Why does metal A give photoelectrons with more kinetic energy than metal B?

Maximum kinetic energy is $K_{max} = h\nu - \phi$ . Metal A has the smaller threshold frequency $(\nu_1 < \nu_2)$ , so its work function $\phi_A = h\nu_1$ is smaller. At the same incident frequency the two metals absorb the same photon energy $h\nu$ , but A spends less of it escaping the surface, leaving more as kinetic energy. So $K_{max}$ is larger for A by exactly $\phi_B - \phi_A = h(\nu_2 - \nu_1)$ .

Does the maximum kinetic energy depend on the intensity of the light?