A) \[+\text{ }3.4eV\]
B) \[+\text{ }6.8\text{ }eV\]
C) \[~-\text{ }13.6\text{ }eV\]
D) \[+\text{ }13.6\text{ }eV\]
Correct Answer: A
Solution :
\[\because \]Total energy \[({{E}_{n}})=KE\,+PE\] In first excited state \[=\frac{1}{2}m{{v}^{2}}+\left[ -\frac{Z{{e}^{2}}}{r} \right]\] \[=+\frac{1}{2}{{\frac{Ze}{r}}^{2}}-\frac{Z{{e}^{2}}}{r}-3.4\,eV\] \[=-\frac{1}{2}\frac{Z{{e}^{2}}}{r}\] \[\therefore \] \[KE=\frac{1}{2}\frac{Z{{e}^{2}}}{r}=+3.4\,eV\]
You need to login to perform this action.
You will be redirected in
3 sec
