In a hydrogen-like ion, the energy difference between the second excited state and the ground state is 108.8 eV. The atomic number of the ion is:
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Question

In a hydrogen-like ion, the energy difference between the second excited state and the ground state is 108.8 eV. The atomic number of the ion is:

G1 Physics · Accepted Answer

The second excited state corresponds to n=3, while the ground state is n=1. For a hydrogen-like ion, E n E_n E n ​ = -13.6Z²/n² eV. Therefore, ΔE = 13.6Z²(1 - 1/9) = 13.6Z²(8/9). Given ΔE = 108.8 eV: 13.6×(8/9)Z² = 108.8. Since 108.8 = 13.6×8, Z²/9 = 1 ⇒ Z = 3. Thus the physics calculation gives option D. The supplied PDF also prints an 'NTA Ans. (2)' line, which conflicts with its own worked solution; this Excel keeps the worked-solution result.