A) 2.3 eV
B) 0.025 eV
C) 10 eV
D) 0.23 eV
Correct Answer: A
Solution :
\[E=\frac{hc}{\lambda }J=\frac{hc}{\lambda \times 1.6\times {{10}^{-19}}}eV\] \[\Rightarrow \]\[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5400\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV\] \[\Rightarrow \]\[E=2.3\,\,eV\]
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