Dual Nature Of Light Question 95
Question: Photoelectric emission is observed from a metallic surface for frequencies $ {\nu_{1}} $ and $ {\nu_{2}} $ of the incident light rays $ ({\nu_{1}}>{\nu_{2}}) $ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $ 1:k $ , then the threshold frequency of the metallic surface is [EAMCET (Engg.) 2001]
Options:
A) $ \frac{{\nu_{1}}-{\nu_{2}}}{k-1} $
B) $ \frac{k{\nu_{1}}-{\nu_{2}}}{k-1} $
C) $ \frac{k{\nu_{2}}-{\nu_{1}}}{k-1} $
D) $ \frac{{\nu_{2}}-{\nu_{1}}}{k} $
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Answer:
Correct Answer: B
Solution:
By using $ h\nu -h{\nu_{0}}={K_{\max }} $
$ \Rightarrow h({\nu_{1}}-{\nu_{0}})=K_{1} $ ?.(i) And $ h({\nu_{2}}-{\nu_{0}})=K_{2} $ ?.(ii)
$ \Rightarrow \frac{{\nu_{1}}-{\nu_{0}}}{{\nu_{2}}-{\nu_{0}}}=\frac{K_{1}}{K_{2}}=\frac{1}{K} $ , Hence $ {\nu_{0}}=\frac{K{\nu_{1}}-{\nu_{2}}}{K-1} $ .
BETA
