A) 2.49 eV
B) 4.49 eV
C) 0.49 eV
D) 5.49 eV
Correct Answer: C
Solution :
\[K{{E}_{\max }}=hv-\phi \] where hv = energy of incident photon, \[\phi \] = work function \[K{{E}_{\max }}=6.6\times {{10}^{-34}}\times 6\times {{10}^{14}}-2\times 1.6\times {{10}^{-19}}\] \[=3.96\times {{10}^{-19}}-3.2\times {{10}^{-19}}\] \[=\frac{0.76\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=0.475eV\]You need to login to perform this action.
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