A) 540 nm
B) 400 nm
C) 310 nm
D) 220 nm
Correct Answer: C
Solution :
The minimum energy of incident radiations, required to eject the electrons from metallic surface is defined as work function of that surface. \[{{W}_{0}}=h{{v}_{0}}=\frac{hc}{{{\lambda }_{0}}}\] where\[{{\lambda }_{0}}=\]threshold wavelength. Work function in electron volt\[{{W}_{0}}(eV)\] \[{{W}_{0}}=\frac{12400}{{{\lambda }_{0}}(\overset{o}{\mathop{\text{A}}}\,)}eV\] \[\therefore \] \[4.0=\frac{12400}{{{\lambda }_{0}}}\] \[\Rightarrow \] \[{{\lambda }_{0}}=310\,nm\]
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