A) \[v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)\]
B) \[{{v}_{1}}+{{v}_{2}}={{\left[ \frac{2h}{m}\left( {{f}_{1}}+{{f}_{2}} \right) \right]}^{1/2}}\]
C) \[v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}\left( {{f}_{1}}+{{f}_{2}} \right)\]
D) \[{{v}_{1}}-{{v}_{2}}={{\left[ \frac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right) \right]}^{1/2}}\]
Correct Answer: A
Solution :
[a] For one photocathode \[h{{f}_{1}}-W=\frac{1}{2}mv_{1}^{2}\] ....(i) For another photo cathode \[h{{f}_{2}}-W=\frac{1}{2}mv_{2}^{2}\] ...(ii) Subtracting (ii) from (i) we get \[\left( h{{f}_{1}}-W \right)\left( h{{f}_{2}}-W \right)=\frac{1}{2}mv_{1}^{2}-\frac{1}{2}mv_{2}^{2}\] \[\therefore h\left( {{f}_{1}}-{{f}_{2}} \right)=\frac{m}{2}\left( v_{1}^{2}-v_{2}^{2} \right)\] \[\therefore v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)\]
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