Dual Nature Of Light Question 410
Question: Photoelectric emission is observed from a metallic surface for frequencies $ v_{1} $ and $ v_{2} $ of the incident light rays $ ( v_{1}>v_{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
Options:
A) $ \frac{v_{1}-v_{2}}{k-1} $
B) $ \frac{kv_{1}-v_{2}}{k-1} $
C) $ \frac{kv_{2}-v_{1}}{k-1} $
D) $ \frac{v_{2}-v_{1}}{k-1} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] By using
$ \Rightarrow h( v_{1}-v_{0} )=K_{1} $ ….(i) $ \text{And }h( v_{2}-v_{0} )=K_{2} $ ….(ii)
$ \Rightarrow \frac{v_{1}-v_{0}}{v_{2}-v_{0}}=\frac{K_{1}}{K_{2}}=\frac{1}{K’} $ $ \text{Hence }v_{0}=\frac{kv_{1}-v_{2}}{K-1}. $
BETA
