A) 7 eV
B) 3.2 eV
C) 1 eV
D) 1.6 eV
Correct Answer: D
Solution :
In the visible light \[{{\lambda }_{\max }}=7800\overset{\text{o}}{\mathop{\text{A}}}\,\] \[E=\frac{hc}{\lambda }=\frac{12375}{7800}eV=1.6\,eV\]
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