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Q.

If the 1^st ionization energy of H atom is 13.6eV, then the 2^nd ionization energy of He atom is

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a

27 .2 eV

b

40.8 eV

c

54.4 eV

d

108.8 eV

answer is C.

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Detailed Solution

$\begin{array}{l} E_{H} = \frac{- 13 .6}{n^{2}}; {(E_{1})}_{H} = - 13 .6 eV \\ {(E_{1})}_{{He}^{+}} = - 13 .6 \frac{Z^{2}}{n^{2}} = - 13 .6 \times \frac{2^{2}}{1} = - 54 .4 eV \end{array}$
The second $I . E . = E_{\infty} - {(E_{1})}_{{He}^{+}}$
$= 0 - (- 54 .4) eV = 54 .4 eV$

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