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KCET · Physics · Nuclear Physics

The ionisation energy of an electron in the ground state of helium atom is \(24.6 \mathrm{eV}\). The energy required to remove both the electron is

  1. A \(51.8 \mathrm{eV}\)
  2. B \(79 \mathrm{eV}\)
  3. C \(38.2 \mathrm{eV}\)
  4. D \(49.2 \mathrm{eV}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(79 \mathrm{eV}\)

Step-by-step Solution

Detailed explanation

Ionisation energy in ground state \(=24.6 \mathrm{eV}\)
Energy required to remove \(2^{n}\) electron from \(\mathrm{He}^{2+}\)
Ionisation energy in ground state \(=24.6 \mathrm{eV}\)
Energy required to remove \(2^{n}\) electron from \(\mathrm{He}^{2+}\)
\(=Z^{2}(13.6) \mathrm{eV}=(2)^{2}(13.6)=54.4 \mathrm{eV}\)
\(=Z^{2}(13.6) \mathrm{eV}=(2)^{2}(13.6)=54.4 \mathrm{eV}\)
So, total energy required \(=24.6+54.4=79 \mathrm{eV}\)
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