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