A) 0.1 eV
B) 2 eV,
C) 0.581 eV
D) 1.581 eV
Correct Answer: C
Solution :
Energy of incident radiation \[E=\frac{hc}{\lambda }\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}J\] \[=\frac{6.6\times 3\times {{10}^{-19}}}{5}=3\times 96\times {{10}^{-19}}J\] \[=\frac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=2.48\,eV\] \[\therefore \]K.E. of emitted photo electrons \[=E-W\] \[=2.48-1.90=0.58\text{ }eV\]
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