The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

Dual Nature of Radiation PYQ — JEE Main 2021 · 25 Feb · Shift 2 · Q67 with Solution

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Correct answer: D

For a given photosensitive surface,

$eV_s = h\nu - \phi = \frac{hc}{\lambda} - \phi$

If we write this in electron-volts, then

$V_s = \frac{1240}{\lambda(\text{nm})} - \frac{\phi}{e}$

For the same surface, the work function term is constant.

For $\lambda_1 = 491\,\text{nm}$ and stopping potential $V_1 = 0.710\,\text{V}$ :

$V_1 = \frac{1240}{491} - \frac{\phi}{e}$

For the new wavelength $\lambda_2$ and stopping potential $V_2 = 1.43\,\text{V}$ :

$V_2 = \frac{1240}{\lambda_2} - \frac{\phi}{e}$

Subtract the first from the second:

$V_2 - V_1 = 1240\left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$

$1.43 - 0.710 = 1240\left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$

$0.720 = 1240\left(\frac{1}{\lambda_2} - \frac{1}{491}\right)$

So,

$\frac{1}{\lambda_2} - \frac{1}{491} = \frac{0.720}{1240}$

$\frac{1}{\lambda_2} = \frac{1}{491} + \frac{0.720}{1240}$

Now compute:

$\frac{1}{491} \approx 0.0020367$

$\frac{0.720}{1240} \approx 0.0005806$

Thus,

$\frac{1}{\lambda_2} \approx 0.0020367 + 0.0005806 = 0.0026173$

Therefore,

$\lambda_2 \approx \frac{1}{0.0026173} \approx 382\,\text{nm}$

The new wavelength is

$\boxed{382\,\text{nm}}$

So the correct option is D.