Dual Nature Of Light Question 386
Question: Two identical photocathodes receive light of frequencies $ f_{1} $ and $ f_{2} $ . If the velocities of the photo electrons (of mass m) coming out are respectively
Options:
A) $ v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}( f_{1}-f_{2} ) $
B) $ v_{1}+v_{2}={{[ \frac{2h}{m}( f_{1}+f_{2} ) ]}^{1/2}} $
C) $ v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}( f_{1}+f_{2} ) $
D) $ v_{1}-v_{2}={{[ \frac{2h}{m}( f_{1}-f_{2} ) ]}^{1/2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] For one photocathode $ hf_{1}-W=\frac{1}{2}mv_{1}^{2} $ ….(i) For another photo cathode $ hf_{2}-W=\frac{1}{2}mv_{2}^{2} $ …(ii) Subtracting (ii) from (i) we get $ ( hf_{1}-W )( hf_{2}-W )=\frac{1}{2}mv_{1}^{2}-\frac{1}{2}mv_{2}^{2} $
$ \therefore h( f_{1}-f_{2} )=\frac{m}{2}( v_{1}^{2}-v_{2}^{2} ) $
$ \therefore v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}( f_{1}-f_{2} ) $
BETA
