Two electrons are moving in orbits of two hydrogen like atoms with speeds $$3\times 10^{5} m/s \text{ and } 2.5\times 10^{5} m/s$$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are ____ respectively.

We need to find the energy states (quantum numbers) of electrons in two hydrogen-like atoms given their speeds and that the orbital radii are nearly the same.

Key formulas for hydrogen-like atoms:

Speed of electron in nth orbit: $$v_n = \frac{Ze^2}{2\epsilon_0 n h}$$, which gives $$v \propto \frac{Z}{n}$$

Radius of nth orbit: $$r_n = \frac{n^2 a_0}{Z}$$, which gives $$r \propto \frac{n^2}{Z}$$

$$v_1 = 3 \times 10^5$$ m/s, $$v_2 = 2.5 \times 10^5$$ m/s, and $$r_1 \approx r_2$$

Since radii are approximately equal:

$$\frac{n_1^2}{Z_1} = \frac{n_2^2}{Z_2}$$

$$\frac{Z_1}{Z_2} = \frac{n_1^2}{n_2^2} \quad \cdots (1)$$

From the velocity relation:

$$\frac{v_1}{v_2} = \frac{Z_1/n_1}{Z_2/n_2} = \frac{Z_1 n_2}{Z_2 n_1}$$

Substituting from equation (1): $$\frac{Z_1}{Z_2} = \frac{n_1^2}{n_2^2}$$

$$\frac{v_1}{v_2} = \frac{n_1^2}{n_2^2} \times \frac{n_2}{n_1} = \frac{n_1}{n_2}$$

$$\frac{3 \times 10^5}{2.5 \times 10^5} = \frac{n_1}{n_2}$$

$$\frac{n_1}{n_2} = \frac{3}{2.5} = \frac{6}{5}$$

So $$n_1 = 6$$ and $$n_2 = 5$$ (taking the simplest integer ratio).

Therefore, the possible energy states are Option 3: 6 and 5.