A) 540 nm
B) 400 nm
C) 310nm
D) 220 nm
Correct Answer: C
Solution :
The work function is given by\[\frac{hc}{{{\lambda }_{0}}}=\phi \] where,\[{{\lambda }_{0}}\]is the threshold wavelength, \[\Rightarrow \] \[{{\lambda }_{0}}=\frac{hc}{\phi }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4\times 1.6\times {{10}^{-19}}}\] \[(\because 1\,EV=1.6\times {{10}^{-19}}J)\] \[=310nm\]
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