A) 1.89 eV
B) 2.89 eV
C) 3.89 eV
D) 4.89 eV
Correct Answer: A
Solution :
Energy of photon \[E=hv=\frac{hc}{\lambda }\] \[=\frac{6.625\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6560\times {{10}^{-10}}}J\] \[=\frac{6.625\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6560\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV\] \[=\frac{6.625\times 3}{6.560\times 1.6}=1.89eV\]
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