A) \[6.8\,\,eV\]
B) \[-13.6\,\,eV\]
C) \[-27.2\,\,eV\]
D) \[-54.4\,\,eV\]
Correct Answer: B
Solution :
\[{{E}_{H{{e}^{+}}}}=\frac{{{Z}^{2}}}{{{n}^{2}}}\times 13.6\,\,eV\] For \[H{{e}^{+}}\] ion \[Z=2\] and for first excited state\[n=2\] \[\therefore \] \[{{E}_{H{{e}^{+}}}}=-\frac{{{2}^{2}}}{{{2}^{2}}}\times 13.6\,\,eV\] \[=-13.6\,\,eV\]
You need to login to perform this action.
You will be redirected in
3 sec
