A) 2.3 eV
B) 3.1 eV
C) 4.5 eV
D) None of these
Correct Answer: B
Solution :
The minimum energy required for the emission of photoelectron from a metal is called the work function W of that metal. If\[{{v}_{0}}\]is threshold frequency, then \[W=h{{v}_{0}}\] but frequency \[({{v}_{0}})=\frac{velocity(c)}{wavelength(\lambda )}\] \[\therefore \] \[W=\frac{hc}{\lambda }\] where,\[h\] is Plancks constant and c the speed of light. Given, \[h=6.6\times {{10}^{-34}}Js,c=3\times {{10}^{8}}m/s,\] \[\lambda =400\text{ }nm=400\times {{10}^{-9}}m\] \[\therefore \] \[W=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{400\times {{10}^{-9}}}J\] \[W=4.95\times {{10}^{-19}}J\] \[W=\frac{4.95\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=3.1\,eV\]
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