Question

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

• A. $27.2\, eV$
• B. $40.8\, eV$
• C. $54.4\, eV$
• D. $108.8\, eV$

Answer 1

Answer: (C)

$54.4\, eV$

Answer 2

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