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
- A \(51.8 \mathrm{eV}\)
- B \(79 \mathrm{eV}\)
- C \(38.2 \mathrm{eV}\)
- D \(49.2 \mathrm{eV}\)
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 \(=24.6+54.4=79 \mathrm{eV}\)
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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