In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: ( hc \over e = 1240 nm eV)

Dual Nature of Radiation PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q56 with Solution

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Correct answer: B

Use Einstein’s photoelectric equation

The stopping potential $V_s$ is related to the maximum kinetic energy: $eV_s = h\nu - \phi = \frac{hc}{\lambda} - \phi$

For two wavelengths, the change in stopping potential is: $e\,\Delta V_s = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2}$

Hence, $\Delta V_s = \frac{hc}{e}\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$
Substitute the given values

Initial wavelength: $\lambda_1 = 300\,\text{nm}$ Final wavelength: $\lambda_2 = 400\,\text{nm}$ Given: $\frac{hc}{e} = 1240\,\text{nm V}$

So, $\Delta V_s = 1240\left(\frac{1}{300}-\frac{1}{400}\right)$
Calculate the bracket

$\frac{1}{300}-\frac{1}{400} = \frac{4-3}{1200} = \frac{1}{1200}$

Therefore, $\Delta V_s = 1240 \times \frac{1}{1200} \approx 1.033\,\text{V}$
Closest option

$\Delta V_s \approx 1.0\,\text{V}$
Option check
- A: $0.5\,\text{V}$ ❌
- B: $1.0\,\text{V}$ ✅
- C: $2.0\,\text{V}$ ❌
- D: $1.5\,\text{V}$ ❌

Therefore, the decrease in stopping potential is close to $1.0\,\text{V}$ .