Question

The energy of the hydrogen atom in its ground state (in eV) is \(-x\). The energy of \(He^+\) ion in its fourth orbit (in eV) is:

• A. \(+\dfrac{x}{5}\)
• B. \(-\dfrac{x}{2}\)
• C. \(+4x\)
• D. \(-\dfrac{x}{4}\)

Hint:

For any hydrogen-like species, \[ E_n=-\frac{13.6Z^2}{n^2}\ \text{eV} \] Always remember that energy is directly proportional to \(Z^2\) and inversely proportional to \(n^2\).

Answer 1

The Correct Option is D

Step 1: Energy expression for hydrogen-like species.
The electron energy in a hydrogen-like ion is given by En = -13.6 Z² / n² eV, where Z is the atomic number and n the principal quantum number. For hydrogen, Z=1 and n=1, so E₁ = -13.6 eV. Given E₁ = -x, we have x = 13.6.

Step 2: Computing the energy for He⁺ in the fourth orbit.
For He⁺, Z=2 and n=4. Plugging into the formula: E₄ = -13.6 (2)² / (4)² = -13.6 × 4 / 16 = -13.6 / 4. Since x = 13.6, this simplifies to -x/4.

Step 3: Checking against the choices.
The derived expression -x/4 corresponds to option (4).

Step 4: Final conclusion.
Hence, the energy of the He⁺ ion in its fourth orbit is -x/4.