A) ultra-violet region
B) infra-red region
C) visible region
D) X-ray region
Correct Answer: A
Solution :
[a] \[\phi =6.2eV=6.2\times 1.6\times {{10}^{-19}}J\] \[V=5volt\] \[\frac{hc}{\lambda }-\phi =e{{V}_{0}}\Rightarrow \lambda =\frac{hc}{\phi +e{{V}_{0}}}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\left( 6.2+5 \right)}\] This range lies in ultra violet range.
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