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Two identical photocathodes receive light of frequencies f_{1} and f_{2} . If the velocities of the photoelectron (of mass m) coming out are v_1 and v_2  respectively, then the correct relation among following is

Option: 1

v_{1}-v_{2}=\left [ \frac{2h}{m}(f_{1}-f_{2}) \right ]^{1/2}

 

 

 

 


Option: 2

V_{1}^{2}-V_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})


Option: 3

v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})


Option: 4

v_{1}+v_{2}=\left [ \frac{2h}{m}(f_{1}-f_{2}) \right ]^{1/2}


Answers (1)

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As we learned

 

 

Einstein’s Photoelectric equation can also be written as - 

                                                                                                                      \mathrm{K.E.}=\mathrm{hf}-\Phi

 

hf_{1}-hf_{0}=\frac{1}{2}mv_{1}^{2}

hf_{2}-hf_{0}=\frac{1}{2}mv_{2}^{2}

\Rightarrow h(f_{1}-f_{2})=\frac{1}{2}m(v_{1}^{2}-v_{2}^{2})

\Rightarrow v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})

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