A) 310 nm
B) 620 nm
C) 6200 nm
D) 3100 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, \[h=6.6\times {{10}^{-34}}J-s,\text{ }c=3\times {{10}^{8}}\,m/s\] \[W=2eV=2\times 1.6\times {{10}^{-19}}J\] \[\therefore \]\[\lambda =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times 1.6\times {{10}^{-19}}}\] \[=620\times {{10}^{-9}}m\] \[\Rightarrow \] \[\lambda =620\text{ }nm\]
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