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In a photoelectric experiment, the wavelength of light incident on a metal is changed from 310 nm to 620 nm. The decrease in stopping potential is close to

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1.5 V

2 V

0.5 V

1 V

answer is D.

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Detailed Solution

$\frac{12400}{3100} = W + e V_{1} ....... (1) \frac{12400}{6200} = W + e V_{2} ....... (2) F r o m (1) & (2) V_{1} - V_{2} = 2 V$

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