Dual Nature Of Light Question 391
Question: The photoelectric threshold wavelength of silver is $ 3250\times {{10}^{-10}}m. $ The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $ 2536\times {{10}^{-10,}} $ m is (Given $ h=4.14\times {{10}^{-15}}eV\text{ and }c=3\times 10^{8}m{{s}^{-1}} $ )
Options:
A) $ =0.6\times 10^{6}m{{s}^{-1}} $
B) $ =61\times 10^{3}m{{s}^{-1}} $
C) $ =0.3\times 10^{6}m{{s}^{-1}} $
D) $ =6\times 10^{6}m{{s}^{-1}} $
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Answer:
Correct Answer: A
Solution:
[a] Given, $ {\lambda_{0}}=3250\times {{10}^{-10}}m $ $ \lambda =2536\times {{10}^{-10}}m $ $ \phi =\frac{hc}{{\lambda_{0}}}=\frac{4.14\times {{10}^{-15}}\times 3\times 10^{8}}{3250\times {{10}^{-10}}}=3.82eV $ $ hv=\frac{hc}{\lambda }=\frac{4.14\times {{10}^{-15}}\times 3\times 10^{8}}{2536\times {{10}^{-10}}}=4.89eV $ According to Einstein’s photoelectric equation, $ {K_{\max }}=hv-\phi $ $ K{E_{\max }}=( 4.89-3.82 )eV=1.077eV $ $ \frac{1}{2}mv^{2}=1.077\times 1.6\times {{10}^{-19}} $
$ \Rightarrow v=\sqrt{\frac{2\times 1.077\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}} $
BETA
