First slide

Photoelectric effect and study of photoelectric effect

Question

Light of wavelength 0.6 pm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 $μ$ m from a mercury vapor lamp, the stopping potential is 1.5 V. Then, the work function [in electron volts] of the photocell surface is

Moderate

A

0.75 eV
B

1.5 eV
C

3 eV
D

2.5 eV

Solution

$\begin{array}{l} E - W_{0} = \frac{1}{2} m v^{2} = e V_{s} \\ or \frac{h c}{λ} - W_{0} = e V_{s} \\ Hence, \frac{h c}{0.6 \times 10^{- 6}} - W_{0} = e (0.5) . . . . . . . . (i) \\ and \frac{h c}{0.4 \times 10^{- 6}} - W_{0} = e (1.5) . . . . . . . . (i i) \\ Solving, we get W_{0} = 1.5 eV \end{array}$

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