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