A) 12.4 eV
B) 6.2 eV
C) 100 eV
D) 200 eV
Correct Answer: B
Solution :
Given, \[{{\lambda }_{\max }}=200mn=2\times {{10}^{-7}}m.\] \[\therefore \] Work function \[(W)=\frac{hc}{{{\lambda }_{\max }}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times {{10}^{-7}}}\] \[=9.9\times {{10}^{-19}}J\] \[=6.19eV\] Energy of photon of wavelength\[100\text{ }nm\]. \[E=\frac{hc}{\lambda }=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1\times {{10}^{-7}}}\] \[=19.8\times {{10}^{-19}}J.\] \[=12.37eV\] \[\therefore \] Maximum KE of photoelectron \[=E-W\] \[=12.37-6.19\] \[=6.18eV\] \[\approx 6.2eV\]You need to login to perform this action.
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