question_answer

Radiation of wavelength \[\lambda ,\]is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to\[\frac{3\lambda }{4},\] the speed of the fastest emitted electron will be:- [JEE MAIN - I 3-4-2016]

A) \[=v{{\left( \frac{3}{5} \right)}^{1/2}}\]

B) \[>v{{\left( \frac{4}{3} \right)}^{1/2}}\]

C) \[<v{{\left( \frac{4}{3} \right)}^{1/2}}\]

D) \[=v{{\left( \frac{4}{3} \right)}^{1/2}}\]

Correct Answer: B

Solution :

[b]

(1)

for fastest emitted electrum