Q154 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Physics

In a photo electric experiment, the wavelength of the light incident on the metal is changed from $200 \mathrm{~nm}$ to $400 \mathrm{~nm}$. The decrease in the stopping potential is close to [Use $\mathrm{hc}=1240 \mathrm{eV}-\mathrm{nm}$ where $\mathrm{h}=$ Planck's constant and $\mathrm{c}$ is velocity of light]

  • Correct answer
    A
    3.1 V
  • B
    2.8 V
  • C
    4.2 V
  • D
    1.2 V

Solution

$\mathrm{eV}_{\mathrm{S}}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}$ $$ \mathrm{V}_{\mathrm{S}}=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) $$ So, $\mathrm{V}_{\mathrm{S} 1}=1240\left(\frac{1}{200}-\frac{1}{\lambda_0}\right)$ $$ \mathrm{V}_{\mathrm{S} 2}=1240\left(\frac{1}{400}-\frac{1}{\lambda_0}\right) $$ So, $\mathrm{V}_{\mathrm{S} 1}-\mathrm{V}_{\mathrm{S} 2}=1240\left(\frac{1}{200}-\frac{1}{400}\right)$ $=1240(5-2.5) \times 10^{-3}=3.1 \mathrm{~V}$

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

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