In a photoelectric experiment|NUCLEAR PHYSICS|JEEMAIN 2019

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JEE MAIN 2019

11-01-2019 S2

Question

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 30 nm . The decrease in the stopping potential is close to: $\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{~nm}-\mathrm{v}\right)$

Select the correct option:

A

1.0 V

B

2.0 V

C

1.5 V

D

0.5 V

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$h\nu = \phi + eV_0 \implies \frac{hc}{\lambda_1} = \phi + eV_1$$$$\frac{hc}{\lambda_2} = \phi + eV_2$$$$1240 \left[ \frac{1}{300} - \frac{1}{400} \right] = e(V_1 - V_2)$$$$(V_1 - V_2) = 1.0 \text{ V}$$

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