question_answer

Two identical, photocathodes receive light of frequencies \[{{f}_{1}}\] and \[{{f}_{2}}\]. If the velocities of the photoelectrons (of mass m) coming out are respectively \[{{v}_{1}}\] and \[{{v}_{2}}\], then         [AIEEE 2003]

A) \[{{v}_{1}}^{2}-{{v}_{1}}^{2}=\frac{2\,h}{m}({{f}_{1}}-{{f}_{2}})\]

B) \[{{v}_{1}}+{{v}_{2}}={{\left[ \frac{2\,h}{m}({{f}_{1}}+{{f}_{2}}) \right]}^{1/2}}\]

C) \[{{v}_{1}}^{2}+{{v}_{2}}^{2}=\frac{2\,h}{m}({{f}_{1}}+{{f}_{2}})\]

D) \[{{v}_{1}}-{{v}_{2}}={{\left[ \frac{2\,h}{m}({{f}_{1}}-{{f}_{2}} \right]}^{1/2}}\]    

Correct Answer: A

Solution :

[a] The Einsteins photo electric equation is

Hence,