A) \[310\text{ }nm\]
B) \[620\text{ }nm\]
C) \[6200\text{ }nm\]
D) \[3100\text{ }nm\]
Correct Answer: B
Solution :
The minimum energy required for the emission of photoelectrons from a metal is called the work function (W) of that metal, and the corresponding wavelength is the threshold wavelength. \[\therefore \] \[W=\frac{hc}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{hc}{W}\] Given, \[h=6.6\times {{10}^{-34}}J-s,\,\,\,c=3\times {{10}^{8}}m/s\] \[W=2eV=2\times 1.6\times {{10}^{-19}}J\] \[\therefore \] \[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times 1.6\times {{10}^{-19}}}\] \[=620\times {{10}^{-9}}m\] \[\Rightarrow \] s\[\lambda =620nm\]
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