Question

Find out the correct energy for the ground state or energy transition. (Symbols have usual meaning and \(n \to m\) gives the transition)

• A. \( \mathrm{H}\;(-6.8\ \text{eV}) \)
• B. \( \mathrm{Li^{2+}}\;(-13.6\ \text{eV}) \)
• C. \( \mathrm{He^+}\;(2 \to 1)\;(40.8\ \text{eV}) \)
• D. \( \mathrm{Be^{3+}}\;(2 \to 1)\;(+13.6\ \text{eV}) \)

Hint:

For hydrogen-like ions, energy scales as \(Z^2\). Always check both magnitude and sign for ground-state energies and transitions.

Answer 1

The Correct Option is C

To determine the correct energy values for the ground state or energy transitions of hydrogen-like ions, it is crucial to understand the formula for calculating energy levels in atoms. The energy of an electron in a hydrogen-like ion is given by:

E_n = − (13.6 × Z²) / n² eV

where:

• E_n is the energy of the n-th level.
• Z is the atomic number (for He⁺, Z = 2).
• n is the principal quantum number.

The energy difference for a transition from level n to m is calculated using:

ΔE = E_n − E_m = 13.6 × Z² ( 1 / m² − 1 / n² ) eV

Let us now solve for the transition 2 → 1 for He⁺:

1. For He⁺, the atomic number is Z = 2.
2. Substitute n = 2 and m = 1 into the formula:

ΔE = 13.6 × 2² ( 1 / 1² − 1 / 2² )

Simplifying:

ΔE = 13.6 × 4 ( 1 − 1 / 4 )

Further simplification gives:

ΔE = 13.6 × 4 × 3 / 4 = 40.8 eV

This confirms that the option He⁺ (2 → 1) (40.8 eV) is correct.

Why the other options are incorrect:

• H (−6.8 eV): The ground state energy of hydrogen is −13.6 eV, not −6.8 eV.
• Li²⁺ (−13.6 eV): For Li²⁺ (Z = 3), the ground state energy is:
  E₁ = −13.6 × 3² = −122.4 eV.
• Be³⁺ (2 → 1) (+13.6 eV): The sign is incorrect because energy is released (negative) during a downward transition. Also, the correct calculation would be:
  ΔE = 13.6 × 4² ( 1 − 1 / 4 ) , which gives a much larger value than stated.

Conclusion:
The correct answer is He⁺ (2 → 1) = 40.8 eV, which matches the calculated transition energy.