Q136 / 150 · MHT CET 2021 (22 Sep Shift 2) · Physics

The light of wavelength ' $\lambda$ ' is incident on the surface of metal of work function $\phi$ and emits the electron, The maximum velocity of electron emitted is [ $\mathrm{m}=$ mass of electron and $\mathrm{h}=$ Planck's constant, $c=$ velocity of light]

  • A
    $\left[\frac{2(\mathrm{hc}-\lambda)^{\frac{1}{2}}}{\mathrm{~m} \lambda}\right]$
  • B
    $\left[\frac{2(\mathrm{hc}-\phi) \lambda}{\mathrm{mc}}\right]$
  • C
    $\left[\frac{2(\mathrm{hc}-\lambda)}{\mathrm{m} \lambda}\right]$
  • Correct answer
    D
    $\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$

Solution

We have $\frac{1}{2} \mathrm{mv}_{\max }^2=\frac{\mathrm{hc}}{\lambda}-\phi=\frac{\mathrm{hc}-\phi \lambda}{\lambda}$ $$ \begin{aligned} & \therefore \mathrm{V}_{\max }^2=\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda} \\ & \therefore \mathrm{V}_{\max }=\sqrt{\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda}} \end{aligned} $$

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

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