A) \[5.5\,\,eV\]
B) \[4\,\,eV\]
C) \[5\,\,eV\]
D) \[4.5\,\,eV\]
Correct Answer: D
Solution :
From Ejection photoelectron \[\because hv=h{{v}_{o}}+K.E.\] \[\frac{hv}{x}=\phi +\text{Stopping}\,\,\text{pot}\] \[\therefore \phi =\frac{hc}{x}-S.P.\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{250\times {{10}^{-9}}\times 1.6\times {{10}^{-19}}}-0.5\times 1.6\times {{10}^{-10}}\] \[=\frac{6.6\times 3}{250\times 1.6}\times {{10}^{2}}-0.5\] \[=4.95-0.5\] \[\tilde{\ }4.5eV\]
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