JEE Physics 2025 Question with Solution

Given: We need the ratio of the radius of the $5^{\text{th}}$ orbit for $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$ in the Bohr model.

Find: $\dfrac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}}$

In the Bohr model, the radius of the $n^{\text{th}}$ orbit for a hydrogen-like atom is

$$r_n = a_0 \frac{n^2}{Z}$$

where $a_0$ is the Bohr radius and $Z$ is the atomic number.

For $\mathrm{Li}^{2+}$, $Z = 3$, so

$$r_5 = \frac{5^2}{3} a_0$$

For $\mathrm{He}^{+}$, $Z = 2$, so

$$r_5 = \frac{5^2}{2} a_0$$

Now take the ratio:

$$\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{\frac{25}{3}a_0}{\frac{25}{2}a_0} = \frac{2}{3}$$

Therefore, the ratio is $\frac{2}{3}$. The correct option is D.

Given: The orbit number is the same, $n = 5$, for both ions.

Find: The ratio of radii for $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$.

Concept Used: For hydrogen-like atoms,

$$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2 Z} = a_0 \frac{n^2}{Z}$$

Thus, for the same orbit, radius is inversely proportional to atomic number:

$$r_n \propto \frac{1}{Z}$$

So,

$$\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}} = \frac{Z_{\mathrm{He}}}{Z_{\mathrm{Li}}} = \frac{2}{3}$$

This matches the computed values

$$r_{5,\mathrm{Li}^{2+}} = a_0 \frac{25}{3}, \qquad r_{5,\mathrm{He}^{+}} = a_0 \frac{25}{2}$$

Hence,

$$\frac{r_{5,\mathrm{Li}^{2+}}}{r_{5,\mathrm{He}^{+}}} = \frac{\frac{25}{3}}{\frac{25}{2}} = \frac{2}{3}$$

Therefore, the correct option is D.