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  4. If the first ionization energy of H atom is 13.6 eV, then the 2nd ionization energy of He atom is

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Q. If the first ionization energy of $H$ atom is $13.6\, eV$, then the $2^{nd}$ ionization energy of He atom is

WBJEEWBJEE 2012Structure of Atom

Solution:

Second ionisation energy $=13.6 \times \frac{2^{2}}{1^{2}}$
$=54.4\, eV$