Question

Light of wavelength $0.6\mu m$ from a sodium lamp falls on a photo cell and causes the emission of photoelectrons for which the stopping potential is 0.5V with light of wavelength $0.4\mu m$ from a sodium lamp, the stopping potential is 1.5V, with this data the value of $\frac{h}{e}$is:

• A. $4\times {10}^{-19}V.s$
• B. $0.25\times {10}^{-15}V.s$
• C. $4\times {10}^{-15}V.s$
• D. $4\times {10}^{-8}V.s$

Answer 1

The correct option is D

$4\times {10}^{-8}V.s$

Equation of photo electric effect
$W=W_0+e{V}_0$
$\frac{hc}{0.6}=W_0+e\times 0.5......\left(1\right)\frac{hc}{0.4}=W_0+1.5e.....\left(2\right)$
(2)-(1)
$hc[\frac{1}{0.4}-\frac{1}{0.6}]=ehc=\frac{0.2}{0.24}=e\frac{hc}{12}=e\frac{h}{e}=\frac{12}{c}=\frac{12}{3\times {10}^8}\frac{h}{e}=4\times {10}^{-8}V-s$