A) \[\frac{13.6}{{{n}^{4}}}eV\]
B) \[\frac{13.6}{{{n}^{3}}}eV\]
C) \[\frac{13.6}{{{n}^{2}}}eV\]
D) \[\frac{13.6}{n}eV\]
Correct Answer: C
Solution :
According to Bohr's, for hydrogen atom\[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\]where, n = principal quantum number
You need to login to perform this action.
You will be redirected in
3 sec
