A) -13.6 eV
B) -27.2 eV
C) -54.4 eV
D) -6.8 eV
Correct Answer: A
Solution :
Energy E of an atom with principal quantum number n is given by \[E=\frac{-13.6}{{{n}^{2}}}{{Z}^{2}}\]for first excited state\[n=2\]and for \[H{{e}^{+}}\,Z=2\] \[\Rightarrow \]\[E=\frac{-13.6\times {{(2)}^{2}}}{{{(2)}^{2}}}=-13.6\,eV\]You need to login to perform this action.
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