A) 15.51 eV
B) 2.91 eV
C) 13.6 eV
D) 1.91 eV
Correct Answer: D
Solution :
The energy of incident photon in joules is \[=\frac{hc}{\lambda }=\frac{6.62\times {{10}^{-34}}\times 3\times {{10}^{-8}}}{800\times {{10}^{-10}}}\] \[=2.4825\times {{10}^{-18\,}}\text{J}\] The energy of the photon in eV is \[\frac{2.4825\times {{10}^{-18}}}{1.6\times {{10}^{-19}}}=15.51\,eV\] The minimum energy needed to ionize the atom = 13.6 eV Hence, KE of the emitted electron =15.51-13.6 =1.91eV
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