A) \[\frac{{{V}_{0}}}{2}\]
B) \[2{{V}_{0}}\]
C) \[{{V}_{0}}+\frac{hc}{2e{{\lambda }_{0}}}\]
D) \[{{V}_{0}}-\frac{hc}{2e{{\lambda }_{0}}}\]
E) \[{{V}_{0}}\]
Correct Answer: D
Solution :
From Einsteins photoelectric equation \[e{{V}_{0}}=\frac{hc}{{{\lambda }_{0}}}-{{W}_{0}}\] \[eV=\frac{hc}{2{{\lambda }_{0}}}-{{W}_{0}}\] Subtracting \[e({{V}_{0}}-V)-\frac{hc}{{{\lambda }_{0}}}\left[ 1-\frac{1}{2} \right]=\frac{hc}{2{{\lambda }_{0}}}\] Or \[V={{V}_{0}}-\frac{hc}{2e{{\lambda }_{0}}}\]
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