WBJEE · Chemistry · Structure of Atom
If the first ionization energy of \(\mathrm{H}\) atom is \(13.6 \mathrm{eV},\) then the second ionization energy of He atom is
- A \(27.2 \mathrm{eV}\)
- B \(40.8 \mathrm{eV}\)
- C \(54.4 \mathrm{eV}\)
- D \(108.8 \mathrm{eV}\)
Answer & Solution
Correct Answer
(C) \(54.4 \mathrm{eV}\)
Step-by-step Solution
Detailed explanation
Second ionisation energy \(=13.6 \times \frac{2^{2}}{1^{2}}\) \[ =54.4 \mathrm{eV} \]
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