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