A) + 54.4 eV
B) ? 13.6 eV
C) + 27.2 eV
D) Cannot be predicted
Correct Answer: B
Solution :
\[E=-\frac{13.6}{{{n}^{2}}}\,\,for\,\,H{{e}^{+}}\,\,n=1\] \[E=-\frac{13.6}{{{1}^{2}}}=-13.6\,eV\].You need to login to perform this action.
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