Q118 / 150 · MHT CET 2026 (11 April Shift 1) · Physics

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays ($v_1 > v_2$). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?

  • A
    $\dfrac{v_1 - v_2}{k}$
  • B
    $\dfrac{v_1 - v_2}{k - 1}$
  • C
    $\dfrac{kv_1 - v_2}{k - 1}$
  • Correct answer
    D
    $\dfrac{kv_2 - v_1}{k - 1}$

Solution

Using Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons is given by $K = h(v - v_0)$, where $v_0$ is the threshold frequency.

For frequency $v_1$, $K_1 = h(v_1 - v_0)$

For frequency $v_2$, $K_2 = h(v_2 - v_0)$

Given that the ratio of maximum kinetic energies is $k : 1$, we have:

$\dfrac{K_1}{K_2} = \dfrac{h(v_1 - v_0)}{h(v_2 - v_0)} = k$

$\dfrac{v_1 - v_0}{v_2 - v_0} = k$

$v_1 - v_0 = k(v_2 - v_0)$

$v_1 - v_0 = kv_2 - kv_0$

$kv_0 - v_0 = kv_2 - v_1$

$v_0(k - 1) = kv_2 - v_1$

$v_0 = \dfrac{kv_2 - v_1}{k - 1}$

Answer: $\dfrac{kv_2 - v_1}{k - 1}$

Hence, the correct answer is option: 'D'.

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