In a hydrogen like ion, the energy difference between the $2^{\text {nd }}$ excitation energy state and…

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
  1. 4
  2. 2
  3. 1
  4. 3

Solution

$\Delta \mathrm{E}=13.6 \mathrm{z}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
$(13.6) \mathrm{z}^2\left[\frac{1}{1}-\frac{1}{9}\right]=108.8$
$\frac{(13.6)(8)}{9}\left(z^2\right)=108.8$
$\mathrm{z}=3$

Asked in: JEE Main 2025 (07 Apr Shift 1)

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