Q 12-12-032JEE MainJEE Main 2026 (24 Jan, Shift 1)Medium
Two electrons are moving in orbits of two hydrogen like atoms with speeds $3\times10^5\ \text{m/s}$ and $2.5\times10^5\ \text{m/s}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are ______ respectively.
Answer: (D) 6 and 5
For a hydrogen-like atom (Bohr model): $r\propto\dfrac{n^2}{Z}$ and $v\propto\dfrac Zn$.
Equal radii: $\dfrac{n_1^2}{Z_1} = \dfrac{n_2^2}{Z_2}$, so $Z\propto n^2$ and then $v\propto\dfrac{n^2}{n} = n$.
$$\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3}{2.5} = \frac65$$
Among the options only $6$ and $5$ has this ratio.
Solution by Sreeraj P, M.Sc Physics