Question

The energy of \(He^+\) ion in its first excited state is, (The ground state energy for the Hydrogen atom is – 13.6 eV) :

• A. -13.6 eV
• B. -3.4 eV
• C. -54.4 eV
• D. -27.2 eV

Answer 1

The Correct Option is A

To determine the energy of the \(He^+\) ion in its first excited state, we need to understand the concept of energy levels in a hydrogen-like ion.

The formula for the energy of an electron in a hydrogen-like ion is given by:

E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}

where:

• Z is the atomic number of the element (for \(He\), Z = 2).
• n is the principal quantum number (for the first excited state, n = 2).

Let's calculate the energy:

E_2 = -13.6 \times \frac{2^2}{2^2} \text{ eV} = -13.6 \text{ eV}

Therefore, the energy of the \(He^+\) ion in its first excited state is -13.6 \text{ eV}, which is confirmed in the given options.

To understand why other options are incorrect:

• -3.4 \text{ eV} would typically correspond to the second excited state of hydrogen, not \(He^+\).
• -54.4 \text{ eV} might relate to a ground state calculation for a higher Z value or different n value.
• -27.2 \text{ eV} would correspond to the ground state energy of the \(He^+\) ion (n = 1).

The correct answer is therefore -13.6 eV in the first excited state of a \(He^+\) ion.