A) 0.1 eV
B) 2 eV
C) 0 58 eV
D) 1.581 eV
Correct Answer: C
Solution :
From photoelectric equation \[E=hv=\frac{hc}{\lambda }\] \[\therefore \] \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}\] \[=3.96\times {{10}^{-19}}J\] Also, \[1eV=1.6\times {{10}^{-19}}J\] \[\therefore \] \[E=\frac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=2.48\,\,eV\] Hence,\[{{E}_{k}}=2.48-1.90=0.58\,\,eV\]
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