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JEE Main 2019 (11 Jan, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2019 (11 Jan, Shift 2)

Q134JEE Main 2019 · 11 JAN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

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: ( $\frac{hc}{e}$ = 1240 nm eV)

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Answer: (b)

$eV_s = \frac{hc}{\lambda} - \phi \Rightarrow \Delta V = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$
$\Delta V = 1240\left(\frac{1}{300} - \frac{1}{400}\right) = \frac{1240}{1200}$
$\Delta V = 1.03\text{ V}$
So the answer is (b) 1.0 V.

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