A) \[14\times {{10}^{-19}}\ J\]
B) \[2.8\times {{10}^{-19}}\ J\]
C) \[1.4\times {{10}^{-19}}J\]
D) \[1.4\times {{10}^{-19}}\,eV\]
Correct Answer: C
Solution :
\[{{K}_{\max }}=\frac{hc}{\lambda }-{{W}_{0}}=\frac{6.4\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6400\times {{10}^{-10}}}-1.6\times {{10}^{-19}}\] \[=1.4\times {{10}^{-19}}\,J\]
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