A) 8
B) 6
C) 4
D) 2
Correct Answer: B
Solution :
Given, \[W=4.125eV=4.125\times 1.6\times {{10}^{-19}}J.\] If \[{{\lambda }_{0}}\] is the cut off wavelength for this surface \[{{\lambda }_{0}}=\frac{hc}{W}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4.125\times 1.6\times {{10}^{-19}}}\] \[=3000\times {{10}^{-10}}=3000\overset{\text{o}}{\mathop{\text{A}}}\,\]
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