A) 420 nm
B) 350 nm
C) 300 nm
D) 200 nm
Correct Answer: C
Solution :
The work function \[{{W}_{0}}=4.125eV=4.125\times 1.6\times {{10}^{-19}}\]joule The relation for work function is\[{{W}_{0}}=\frac{hc}{\lambda }\](where \[\lambda \] is the cut off wavelength) \[6.6\times {{10}^{-19}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }\] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6.6\times {{10}^{-19}}}\] \[=\frac{19.8\times {{10}^{-26}}}{6.6\times {{10}^{-19}}}=3\times {{10}^{-7}}cm.\] \[=300\times {{10}^{-9}}m=300nm\]
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