A) 2.5 eV
B) 50 V
C) 5.48 eV
D) 7.48 eV
Correct Answer: A
Solution :
Energy of photon \[E=\frac{hc}{\lambda }\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}\] \[=3.96\times {{10}^{-19\,}}J\] \[=\frac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=2.475eV\approx 2.5eV\]
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