Atomic Structure Question 199
Question: A photon of frequency n causes photoelectric emission from a surface with threshold frequency $ n_0 $ .The de Broglie wavelength $ \lambda $ of the photoelectron emitted is given as
Options:
A) $ \Delta n=\frac{h}{2m\lambda } $
B) $ \Delta n=\frac{h}{\lambda } $
C) $ [ \frac{1}{v_0}-\frac{1}{v} ]=\frac{mc^{2}}{h} $
D) $ \lambda =\sqrt{\frac{h}{2m\Delta n}} $
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Answer:
Correct Answer: D
Solution:
- $ E_1=IE+KE $ Or $ E_1= $ threshold E (or) work function +KE $ ( hv=hv_0+\frac{1}{2}mu^{2} ) $ Or $ hn=hn_0+\frac{1}{2}mu^{2} $
$ \frac{1}{2}mu^{2}=h(n-n_0)=h\Delta n $
$ ( \lambda =\frac{h}{mu},\therefore u=\frac{h}{m\lambda } ) $ Substitute the value of u I equation (i) $ \frac{1}{2}m.\frac{h^{2}}{m^{2}{{\lambda }^{2}}}=h\Delta n $
$ \frac{h}{2{{\lambda }^{2}}m}=\Delta n $
$ \therefore $ $ \lambda =\sqrt{\frac{h}{2m\Delta n}} $
