JEE Physics 2025 Question with Solution

Given: The energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \, \text{eV}$.

Find: The atomic number $Z$ of the hydrogen-like ion.

For a hydrogen-like ion, the energy of the $n^{\text{th}}$ level is

$$E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2}$$

The $2^{\text{nd}}$ excitation state corresponds to $n = 3$, and the ground state corresponds to $n = 1$.

So,

$$E_3 = -\frac{13.6 Z^2}{9}$$

and

$$E_1 = -13.6 Z^2$$

The energy difference is

$$\Delta E = E_1 - E_3$$ $$\Delta E = -13.6 Z^2 - \left(-\frac{13.6 Z^2}{9}\right)$$ $$\Delta E = 13.6 Z^2\left(1 - \frac{1}{9}\right)$$ $$\Delta E = 13.6 Z^2 \times \frac{8}{9}$$ $$\Delta E = \frac{108.8 Z^2}{9}$$

Given that $\Delta E = 108.8 \, \text{eV}$,

$$\frac{108.8 Z^2}{9} = 108.8$$ $$Z^2 = 9$$ $$Z = 3$$

Therefore, the atomic number of the ion is $3$. The correct option is D.

Given: $\Delta E = 108.8 \, \text{eV}$.

Find: The atomic number $Z$.

Use the direct energy gap relation for a hydrogen-like ion:

$$\Delta E = 13.6 Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Here, $n_1 = 1$ and $n_2 = 3$, so

$$\Delta E = 13.6 Z^2 \left(1 - \frac{1}{9}\right)$$ $$108.8 = 13.6 Z^2 \times \frac{8}{9}$$

Since $13.6 \times 8 = 108.8$,

$$108.8 = \frac{108.8 Z^2}{9}$$ $$Z^2 = 9$$ $$Z = 3$$

This works quickly because the required transition is directly from $n=3$ to $n=1$, so substituting the two quantum numbers into the level-gap formula avoids calculating each level separately. Therefore, the correct option is D.