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