JEE Physics 2025 Question with Solution

Given: We need to identify the correct statements using Bohr's atomic model for hydrogen-like species.

Find: Which pair among (A), (B), (C), (D) is correct.

For a hydrogen-like atom, the energy of the electron in the $n^{\text{th}}$ orbit is

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

where $Z$ is the atomic number.

Also,

• ground state $\Rightarrow n=1$
• first excited state $\Rightarrow n=2$
• second excited state $\Rightarrow n=3$

Now evaluate the relevant comparisons shown in the solution.

For H atom in ground state:

$$E_{\text{H, ground}} = -13.6 \times \frac{1^2}{1^2} = -13.6 \, \text{eV}$$

For He$^+$ in first excited state:

$$E_{\text{He}^+, \text{1st excited}} = -13.6 \times \frac{2^2}{2^2} = -13.6 \, \text{eV}$$

So statement (A) is true.

For Li$^{2+}$ in second excited state:

$$E_{\text{Li}^{2+}, \text{2nd excited}} = -13.6 \times \frac{3^2}{3^2} = -13.6 \, \text{eV}$$

Thus this is also equal to the ground-state energy of hydrogen, so statement (B) is true.

For He$^+$ in ground state:

$$E_{\text{He}^+, \text{ground}} = -13.6 \times \frac{2^2}{1^2} = -54.4 \, \text{eV}$$

This is not equal to $-13.6 \, \text{eV}$, so statement (C) is false.

For Li$^{2+}$ in ground state:

$$E_{\text{Li}^{2+}, \text{ground}} = -13.6 \times \frac{3^2}{1^2} = -122.4 \, \text{eV}$$

This is not equal to the first excited state energy of He$^+$, so statement (D) is false.

Therefore, only (A) and (B) are correct. The correct option is B.

Given: The question is based on Bohr energy levels for hydrogen-like atoms.

Find: Which statements among (A), (B), (C), (D) are correct.

The key dependence is

$$E_n \propto \frac{Z^2}{n^2}$$

So whenever $\frac{Z^2}{n^2}$ is the same for two cases, their energies are equal.

• For H ground state: $\frac{Z^2}{n^2} = \frac{1^2}{1^2} = 1$
• For He$^+$ first excited state: $\frac{2^2}{2^2} = 1$
• For Li$^{2+}$ second excited state: $\frac{3^2}{3^2} = 1$

Hence these three states have the same energy, namely $-13.6 \, \text{eV}$.

So:

• (A) is correct
• (B) is correct

Now check the remaining two:

• He$^+$ ground state gives $\frac{2^2}{1^2} = 4$, so its energy is much lower, not equal to hydrogen ground state. Hence (C) is false.
• Li$^{2+}$ ground state gives $\frac{3^2}{1^2} = 9$, so it is also not equal to He$^+$ first excited state. Hence (D) is false.

Therefore, the correct option is B, that is (A), (B) only.