A) 1.25eV
B) 2.17eV
C) 4.07eV
D) 3.57eV
Correct Answer: B
Solution :
\[KE=e{{V}_{0}}=\frac{hc}{\lambda }-Q\] \[1.6\times {{10}^{-19}}\times 0.38=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{448\times {{10}^{-9}}}-\phi \] \[\phi =2.17eV\]You need to login to perform this action.
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