A) 310 nm
B) 620 nm
C) 1200 nm
D) 2100 nm
Correct Answer: B
Solution :
The minimum energy required for the emission of photoelectron from a metal is called the work function of that metal. \[W=hv=\frac{hc}{\lambda }\] Where \[v\] is frequency, \[c\] is speed of light, \[\lambda \] is wavelength. Given, \[W=2eV=2\times 1.6\times {{10}^{-19}}\,\,J\], \[h=6.6\times {{10}^{-34}}\,J-s,\,c=3\times {{10}^{8}}\,m/s\] \[\Rightarrow \] \[\lambda =\frac{hc}{W}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times 1.6\times {{10}^{-19}}}\] \[=620\times {{10}^{-9}}\,m=620\,nm\]Note: If wavelength of incident light greater than 620 nm no photoelectrons are emitted.
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