Dual Nature Of Light Question 294
Question: Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000 Å falls on the surface, the velocity of the fastest electron ejected from the surface will be [AMU 1999]
Options:
A) $ 8.4\times 10^{5}m/sec $
B) $ 7.4\times 10^{5}m/sec $
C) $ 6.4\times 10^{5}m/sec $
D) $ 8.4\times 10^{6}m/sec $
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Answer:
Correct Answer: A
Solution:
By using $ E=W_{0}+\frac{1}{2}mv_{\max }^{2} $ where $ E=\frac{12375}{2000}=6.18\ eV $
$ \Rightarrow 6.18\ eV=4.2\ eV+\frac{1}{2}mv_{\max }^{2} $
$ \Rightarrow 1.98\ eV=\frac{1}{2}mv_{\max }^{2} $
Þ $ 1.98\times 1.6\times {{10}^{-19}}=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times v_{\max }^{2} $
$ \Rightarrow {v_{\max }}=8.4\times 10^{5}\ m/s $
BETA
