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