A) ultra violet region
B) infrared region
C) visible region
D) X - ray region
Correct Answer: A
Solution :
From Einstein's photoelectric equation \[hv=h{{v}_{0}}+e{{V}_{0}}\] \[=6.2+5=11.2\,\,eV\] \[\Rightarrow \] \[\frac{hc}{\lambda }=11.2\,eV\]or \[\lambda =\frac{hc}{11.2\,eV}\] \[=\frac{6.6\times {{10}^{-34}}\times 3.0\times {{10}^{8}}}{11.2\times 1.6\times {{10}^{-19}}}\] \[=1.1049\times {{10}^{-7}}\] \[=1104.9\overset{\text{o}}{\mathop{\text{A}}}\,\] This incident radiation lies in ultra violet region,
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