In a hydrogen like ion, the energy difference between the $2^{\text{nd}}$ excitation energy state and ground is 108.8 eV. The atomic number of the ion is

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Modern Physics

In a hydrogen like ion, the energy difference between the

2^{nd}

excitation energy state and ground is 108.8 eV. The atomic number of the ion is

(A)4
(B)2
(C)1
(D)3✓ Correct

Answer: (D)

Solution

The second excited state of a hydrogen-like ion is

n = 3

, so the energy released when the electron returns to the ground state is

Δ E = 13.6 Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}) = 13.6 Z^{2} \cdot \frac{8}{9} eV

. Setting this equal to the given

108.8 eV

gives

Z^{2} = \frac{108.8 \times 9}{13.6 \times 8} = 9

, so the atomic number is

Z = 3

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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