A) \[{{\operatorname{V}}_{1}}-\frac{h}{e}\left( {{f}_{2}}-{{f}_{1}} \right)\]
B) \[{{\operatorname{V}}_{1}}+\frac{h}{e}\left( {{f}_{2}}+{{f}_{1}} \right)\]
C) \[{{\operatorname{V}}_{1}}-\frac{h}{e}\left( {{f}_{2}}+{{f}_{1}} \right)\]
D) \[{{\operatorname{V}}_{1}}+\frac{h}{e}\left( {{f}_{2}}-{{f}_{1}} \right)\]
Correct Answer: C
Solution :
Photoelectric effect equation \[{{\operatorname{hf}}_{1}}={{\phi }_{o}}+{{\operatorname{eV}}_{1}}\] \[{{\phi }_{o}}={{\operatorname{hf}}_{1}}-{{\operatorname{eV}}_{1}}\] ?.(1) In incident light of frequency \[{{\operatorname{f}}_{2}}\]is incident \[{{\operatorname{hf}}_{2}}={{\phi }_{o}}+{{\operatorname{eV}}_{1}}\] \[{{\operatorname{eV}}_{1}}={{\operatorname{hf}}_{2}}+{{\phi }_{\operatorname{o}}}\] Same metal is exposed to radiation hence, work function is same \[{{\operatorname{eV}}_{2}}={{\operatorname{hf}}_{2}}+\left[ {{\operatorname{hf}}_{1}}-{{\operatorname{eV}}_{1}} \right]\because \left[ {{\phi }_{o}}-{{\operatorname{hf}}_{1}}-{{\operatorname{eV}}_{1}} \right]\] \[{{V}_{2}}=\frac{\operatorname{h}}{\operatorname{e}}-\left( {{f}_{2}}+{{f}_{1}} \right)-{{\operatorname{V}}_{1}}\] \[or\,\,\,{{V}_{2}}={{\operatorname{V}}_{1}}-\frac{\operatorname{h}}{\operatorname{e}}\left( {{f}_{2}}+{{f}_{1}} \right)\]
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