Electron Energy States: Speed vs. Quantum Number

For electrons in hydrogen-like atoms, speed ($v_n$) is proportional to $Z/n$ and radius ($r_n$) is proportional to $n^2/Z$. Here, $Z$ is the atomic number and $n$ is the principal quantum number.

• Speed dependency: $v_n \propto \frac{Z}{n}$
• Radius dependency: $r_n \propto \frac{n^2}{Z}$

We are given the speeds of two electrons: $v_1 = 3 \times 10^5 \text{ m/s}$ and $v_2 = 2.5 \times 10^5 \text{ m/s}$. The ratio of their speeds is:

$ \frac{v_1}{v_2} = \frac{3 \times 10^5 \text{ m/s}}{2.5 \times 10^5 \text{ m/s}} = \frac{3}{2.5} = \frac{30}{25} = \frac{6}{5} $

The problem states that the radii of their orbits are nearly the same ($r_1 \approx r_2$). This implies:

$ \frac{r_1}{r_2} \approx 1 \implies \frac{n_1^2/Z_1}{n_2^2/Z_2} \approx 1 \implies \frac{Z_1}{Z_2} \approx \frac{n_1^2}{n_2^2} $

Now, consider the ratio of speeds in terms of atomic numbers ($Z_1, Z_2$) and principal quantum numbers ($n_1, n_2$):

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

Substitute the relationship $\frac{Z_1}{Z_2} \approx \frac{n_1^2}{n_2^2}$ derived from the equal radii condition:

$ \frac{v_1}{v_2} \approx \left( \frac{n_1^2}{n_2^2} \right) \times \frac{n_2}{n_1} = \frac{n_1}{n_2} $

Equating the calculated speed ratio with the derived quantum number ratio:

$ \frac{n_1}{n_2} \approx \frac{6}{5} $

We need to find the pair $(n_1, n_2)$ from the options that matches this ratio. The pair $(6, 5)$ gives $\frac{6}{5}$, which matches our result.

The possible order of energy states (principal quantum numbers) is 6 and 5.