A) \[0.6\times {{10}^{-19}}J\]
B) \[2.6\times {{10}^{-19}}J\]
C) \[1.6\times {{10}^{-19}}J\]
D) \[1.6\,eV\]
Correct Answer: A
Solution :
Here \[{{W}_{0}}=1.5\,eV\] and \[\lambda =6000\overset{o}{\mathop{A}}\,\] putting the given value in equation (i), we get \[\therefore \]\[{{F}_{\max }}=\frac{6.62\times {{10}^{-31}}\times 3\times {{10}^{8}}}{6600\times {{10}^{-10}}}-1.5\times 1.6\times {{10}^{-19}}\] \[{{F}_{\max }}=3\times {{10}^{-19}}-2.40\times {{10}^{-19}}\] \[0.6\times {{10}^{-19}}\]Joule
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