JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

Two electrons are moving in orbits of two hydrogen like atoms with speeds $3 \times 1 0^{5}$ m/s and $2.5 \times 1 0^{5}$ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are_____ respectively.

A
$6$ and $5$
B
$9$ and $8$
C
$8$ and $10$
D
$10$ and $12$

Solution

For a hydrogen-like atom (Bohr model): $v_{n} \propto \frac{Z}{n}$ and $r_{n} \propto \frac{n ^{2}}{Z}$ .

If the two orbits have nearly the same radius, then $\frac{n ^{2}}{Z}$ is the same for both, so $Z \propto n^{2}$ .

Substituting into $v \propto Z / n$ : $v \propto \frac{n ^{2}}{n} = n$ .

Therefore $\frac{n _{1}}{n _{2}} = \frac{v _{1}}{v _{2}} = \frac{3 \times 1 0 ^{5}}{2.5 \times 1 0 ^{5}} = \frac{6}{5}$ .

The simplest integer pair is $n_{1} = 6$ and $n_{2} = 5$ .

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