A) \[2.48\text{ }eV\]
B) \[0.41\text{ }eV\]
C) \[2.07\text{ }eV\]
D) \[0.82\text{ }eV\]
Correct Answer: B
Solution :
Work function \[{{W}_{0}}=3.31\times {{10}^{-19}}J\] Wavelength of incident radiation \[\lambda =5000\times {{10}^{-10}}m\] \[E={{W}_{0}}+KE\] (According to Einstein equation) \[\frac{hc}{\lambda }=3.31\times {{10}^{-19}}+KE\] \[KE=-3.31\times {{10}^{-19}}+\frac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}\] \[=-3.31\times {{10}^{-19}}+\frac{6.62\times 3}{5}\times {{10}^{-19}}\] \[=(-3.31\times 1.324\times 3)\times {{10}^{-19}}\] \[=(3.972-3.31)\times {{10}^{-19}}=0.662\times {{10}^{-19}}J\] \[\Rightarrow \] \[E=\frac{0.662\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=0.41eV\]You need to login to perform this action.
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