A) 4eV
B) 2eV
C) 4.2eV
D) 2.5eV
Correct Answer: A
Solution :
From the graph, threshold frequency \[{{v}_{0}}=10\times {{10}^{14}}Hz\] \[h=\frac{\Delta {{k}_{\max }}}{\Delta v}=\frac{8\times 16\times {{10}^{-19}}}{20\times {{10}^{14}}}=6.4\times {{10}^{-34}}J\] Thus, work function, \[W=h{{v}_{0}}=6.4\times {{10}^{-34}}\times 10\times {{10}^{14}}\] \[=4\times 1.6\times {{10}^{-19}}=4\,eV\]
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