Question

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

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

Answer 1

Answer: (C)

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