Q145 / 150 · MHT CET 2024 (09 May Shift 1) · Physics

Two identical photocathodes receive light of frequencies ' $\mathrm{n}_1$ ' and ' $\mathrm{n}_2$ '. If the velocities of the emitted photoelectrons of mass ' $m$ ' are ' $V_1$ ' and ' $\mathrm{V}_2$ ' respectively, then ( $\mathrm{h}=$ Planck's constant )

  • A
    $\mathrm{V}_1+\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)\right]^{1 / 2}$
  • B
    $\mathrm{V}_1-\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)\right]^{1 / 2}$
  • C
    $\mathrm{V}_1^2+\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)$
  • Correct answer
    D
    $\mathrm{V}_1^2-\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)$

Solution

$\begin{aligned} & \text { K.E. } ._1=h n_1-\phi \\ & \text { K.E. } 2=\mathrm{hn}_2-\phi \\ & \therefore \quad \text { K.E. } 1 \text {-K.E. } 2=\text { hn }_1-\mathrm{hn}_2 \\ & \therefore \quad \frac{\mathrm{~m}\left(\mathrm{~V}_1^2-\mathrm{V}_2{ }_2\right)}{2}=\mathrm{h}\left(\mathrm{n}_1-\mathrm{n}_2\right) \\ & \therefore \quad \mathrm{V}_1^2-\mathrm{V}^2{ }_2=2 \mathrm{~h} \frac{\left(\mathrm{n}_1-\mathrm{n}_2\right)}{\mathrm{m}} \end{aligned}$

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

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