A) \[{{v}_{1}}-{{v}_{2}}={{\left[ \frac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right) \right]}^{1/2}}\]
B) \[v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)\]
C) \[{{v}_{1}}+{{v}_{2}}={{\left[ \frac{2h}{m}\left( {{f}_{1}}+{{f}_{2}} \right) \right]}^{1/2}}\]
D) \[v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}\left( {{f}_{1}}+{{f}_{2}} \right)\]
Correct Answer: B
Solution :
Using Einstein photoelectric equation \[E={{W}_{0}}+{{K}_{\max }}\] \[h{{f}_{1}}={{W}_{0}}+\frac{1}{2}mv_{1}^{2}\] ?..(i) \[h{{f}_{2}}={{W}_{0}}+\frac{1}{2}mv_{2}^{2}\] ?..(ii) \[\Rightarrow h({{f}_{1}}-{{f}_{2}})=\frac{1}{2}m(v_{1}^{2}-v_{2}^{2})\]\[\Rightarrow (v_{1}^{2}-v_{2}^{2})=\frac{2h}{m}({{f}_{1}}-{{f}_{2}})\]
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