Dual Nature Of Light Question 93
Question: In a photoemissive cell with executing wavelength $ \lambda $ , the fastest electron has speed v. If the exciting wavelength is changed to $ 3\lambda /4 $ , the speed of the fastest emitted electron will be [CBSE PMT 1998]
Options:
A) $ v\ {{(3/4)}^{1/2}} $
B) $ v\ {{(4/3)}^{1/2}} $
C) Less than $ v\ {{(4/3)}^{1/2}} $
D) Greater than $ v\ {{(4/3)}^{1/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ h\nu -W_{0}=\frac{1}{2}mv_{\max }^{2}\Rightarrow \frac{hc}{\lambda }-\frac{hc}{{\lambda_{0}}}=\frac{1}{2}mv_{\max }^{2} $
$ \Rightarrow hc( \frac{{\lambda_{0}}-\lambda }{\lambda {\lambda_{0}}} )=\frac{1}{2}mv_{\max }^{2} $
$ \Rightarrow {v_{\max }}=\sqrt{\frac{2hc}{m}( \frac{{\lambda_{0}}-\lambda }{\lambda {\lambda_{0}}} )} $ When wavelength is $ \lambda $ and velocity is v, then $ v=\sqrt{\frac{2hc}{m}( \frac{{\lambda_{0}}-\lambda }{\lambda {\lambda_{0}}} )} $ ?. (i) When wavelength is $ \frac{3\lambda }{4} $ and velocity is v? then $ v’=\sqrt{\frac{2hc}{m}[ \frac{{\lambda_{0}}-(3\lambda /4)}{(3\lambda /4)\times {\lambda_{0}}} ]} $ ?.(ii) Divide equation (ii) by (i), we get $ \frac{v’}{v}=\sqrt{\frac{[{\lambda_{0}}-(3\lambda /4)]}{\frac{3}{4}\lambda {\lambda_{0}}}\times \frac{\lambda {\lambda_{0}}}{{\lambda_{0}}-\lambda }} $ $ v’=v{{( \frac{4}{3} )}^{1/2}}\sqrt{\frac{[{\lambda_{0}}-(3\lambda /4)]}{{\lambda_{0}}-\lambda }} $ i.e. $ v’>v{{( \frac{4}{3} )}^{1/2}} $
BETA
