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.
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