Dual Nature Of Light Question 360
Question: Radiation of wavelength $ \lambda $ , is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to $ \frac{3\lambda }{4} $ , the speed of the fastest emitted electron will be:
Options:
A) $ =v{{( \frac{4}{3} )}^{\frac{1}{2}}} $
B) $ =v{{( \frac{3}{4} )}^{\frac{1}{2}}} $
C) $ >v{{( \frac{4}{3} )}^{\frac{1}{2}}} $
D) $ <v{{( \frac{4}{3} )}^{\frac{1}{2}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ hv_{0}^{2}-hv_{0}=\frac{1}{2}mv^{2} $
$ \therefore \frac{4}{3}hv_{0}-hv_{0}=\frac{1}{2}mv{{’}^{2}} $
$ \therefore \frac{v{{’}^{2}}}{v^{2}}=\frac{\frac{4}{3}v-v_{0}}{v-v_{0}}\text{ }\therefore v’=v\sqrt{\frac{\frac{4}{3}v-v_{0}}{v-v_{0}}} $
$ \therefore ,v’>V\sqrt{\frac{4}{3}} $
BETA
