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(A) For a hydrogen-like atom, the velocity of an electron in the $n^{th}$ orbit is given by $v \propto \frac{Z}{n}$.The radius of the $n^{th}$ orbit i

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$6$ and $5$

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(A) For a hydrogen-like atom, the velocity of an electron in the $n^{th}$ orbit is given by $v \propto \frac{Z}{n}$.The radius of the $n^{th}$ orbit is given by $r \propto \frac{n^2}{Z}$.From these two relations, we can write $r \propto \frac{n}{v}$ (since $Z \propto \frac{n}{v} \cdot n = \frac{n^2}{r}$, substituting $Z$ into the velocity relation).Given that the radii are nearly the same $(r_1 \approx r_2)$, we have $\frac{n_1}{v_1} = \frac{n_2}{v_2}$.Therefore, $\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3 	imes 10^5}{2.5 	imes 10^5} = \frac{3}{2.5} = \frac{6}{5}$.Thus, the possible order of energy states ($n_1$ and $n_2$) is $6$ and $5$.

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

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