A) 1.8 eV
B) 2.5 eV
C) 2.1 eV
D) 1.4 eV
Correct Answer: A
Solution :
\[W=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6800\times {{10}^{-10}}}\] \[=2.9\times {{10}^{-19}}J\] \[=\frac{2.9\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=1.8\,eV\]
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