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If the first ionization energy of $\mathrm{H}$ atom is $13.6 \mathrm{eV},$ then the second ionization energy of He atom is
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Verified Answer
The correct answer is:
$54.4 \mathrm{eV}$
Second ionisation energy $=13.6 \times \frac{2^{2}}{1^{2}}$
$$
=54.4 \mathrm{eV}
$$
$$
=54.4 \mathrm{eV}
$$
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