Question

Minimum energy required to take out the only one electron from ground state of He\(^+\) is

• A. \(13.6\ \mathrm{eV}\)
• B. \(54.4\ \mathrm{eV}\)
• C. \(27.2\ \mathrm{eV}\)
• D. \(6.8\ \mathrm{eV}\)

Hint:

Energy levels scale as \(Z^2\) for hydrogen-like ions.

Answer 1

The Correct Option is B

To find the minimum energy required to remove the only electron from the ground state of the hydrogen-like ion \(\mathrm{He}^{+}\), we need to use the formula for the ionization energy of hydrogen-like atoms:

The ionization energy for a hydrogen-like atom is given by:

\(E = 13.6 \times Z^2 \ \mathrm{eV}\)

where:

• \(Z\) is the atomic number.
• \(13.6 \ \mathrm{eV}\) is the ionization energy of hydrogen atom (n = 1 to n = infinity) for a single electron.

For \(\mathrm{He}^{+}\), the atomic number (\(Z\)) is 2. Therefore, substituting into the formula:

\(E = 13.6 \times 2^2 \ \mathrm{eV} = 13.6 \times 4 \ \mathrm{eV} = 54.4 \ \mathrm{eV}\)

The minimum energy required to remove the electron from the ground state of \(\mathrm{He}^{+}\) ion is \(54.4\ \mathrm{eV}\).

Conclusion:

Hence, the correct answer is \(54.4 \ \mathrm{eV}\).

The other options can be ruled out as they do not take into account the correct atomic number for \(\mathrm{He}^{+}\).