Dual Nature Of Light Question 247
Question: The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be [MP PMT 1990]
Options:
A) 10 m/sec
B) $ 1\times 10^{3} $ m/sec
C) $ 1\times 10^{4}m/\sec $ m/sec
D) $ 1\times 10^{6}m/\sec $ m/sec
Show Answer
Answer:
Correct Answer: D
Solution:
$ E=W_{0}+{K_{\max }};\ E=\frac{12375}{3000}=4.125\ eV $
$ \Rightarrow {K_{\max }}=E-W_{0}=4.125\ eV-1\ eV=3.125\ eV $
$ \Rightarrow \frac{1}{2}mv_{\max }^{2}=3.125\times 1.6\times {{10}^{-19}}\ J $
$ \Rightarrow {v_{\max }}=\sqrt{\frac{2\times 3.125\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}} $ $ =1\times 10^{6}m/s $
BETA
