A) 54.4 eV
B) 6.8 eV
C) 13.6eV
D) 24.5 eV
Correct Answer: A
Solution :
: \[{{E}_{n}}=\frac{-13.6}{{{n}^{2}}}{{Z}^{2}}eV\] \[{{(I.P.)}_{H}}={{E}_{\infty }}-{{E}_{1}}=13.6eV\] \[{{(I.P.)}_{H{{e}^{+}}}}={{E}_{\infty }}-{{E}_{1}}=13.6\times {{2}^{2}}=54.4eV\]
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