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 first excited state \[=\frac{1}{2}m{{v}^{2}}+\left[ -\frac{Z{{e}^{2}}}{r} \right]\] |
| \[=+\frac{1}{2}\frac{Z{{e}^{2}}}{r}-\frac{Z{{e}^{2}}}{r}\] |
| \[-3.4\,eV=-\frac{1}{2}\frac{Z\,{{e}^{2}}}{r}\] |
| \[KE=\frac{1}{2}\,\frac{Z{{e}^{2}}}{r}=+3.4\,eV\] |
You need to login to perform this action.
You will be redirected in
3 sec
