JEE Main 2025 Apr 7 Shift 1, Physics Q15: Bohr Model of the Hydrogen Atom

JEE Main2025Apr 7, Shift 1Physics

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

Solution

For a hydrogen-like ion,

$E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} eV$

The 2^nd excited state corresponds to $n = 3$ .

$∴ E_{1} = - 13.6 Z^{2}, E_{3} = - \frac{13.6 Z ^{2}}{9}$

Given,

$E_{3} - E_{1} = 108.8 eV$

$\Rightarrow 13.6 Z^{2} (1 - \frac{1}{9}) = 108.8$

$\Rightarrow 13.6 Z^{2} \cdot \frac{8}{9} = 108.8$

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

$∴ Z = 3$

Hence, the atomic number of the ion is 3.

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