A) +3.4 eV
B) +6.8 eV
C) -13.6 eV
D) +13.6 eV
Correct Answer: A
Solution :
\[\because \]Total energy \[({{E}_{n}})=KE+PE\] In the first excited state \[=\frac{1}{2}m{{v}^{2}}+\left[ -\frac{Z{{e}^{2}}}{r} \right]=+\frac{1}{2}\frac{Z{{e}^{2}}}{r}-\frac{Z{{e}^{2}}}{r}\] \[-3.4\,eV=-\frac{1}{2}\frac{Z{{e}^{2}}}{r}\] \[\therefore \] \[KE=\frac{1}{2}\frac{Z{{e}^{2}}}{2}=+3.4\,eV.\]
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