A) \[-6.8\] eV
B) \[-3.4\] eV
C) \[-1.51\] eV
D) \[-4.53\] Ev
Correct Answer: C
Solution :
\[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] For second excited state \[n=3\], \[{{E}_{3}}=-\frac{13.6}{9}=-1.51\,eV\]You need to login to perform this action.
You will be redirected in
3 sec