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

Given: The electron speeds are v1=3 10^5 m/s and v2=2.5 10^5 m/s. The radii of the two orbits are nearly the same. Find: The possible order of principal energy states. For a hydrogen like atom, the speed of electron in the n^ th orbit is given by vn 1n So, writing the ratio of speeds, v1v2= 3 10^52.5 10^5= 65 Since vn 1n, v1v2= n2n1 Therefore, n2n1= 65 Thus, the possible values of n1 and n2 are 5 and 6 in the same ratio, so the order of energy states corresponding to the given electrons is 6 and 5 respectively. The correct option is C.

Two electrons are moving in orbits of two hydrogen like atoms wit… — Solution | JEE Physics

MCQEasyJEE 2026Bohr's Model & Hydrogen Spectrum

Two electrons are moving in orbits of two hydrogen like atoms with speeds $3\times10^{5}\,m/s$ and $2.5\times10^{5}\,m/s$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are _____ respectively.

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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The electron speeds are $v_1=3\times10^{5}\,m/s$ and $v_2=2.5\times10^{5}\,m/s$ . The radii of the two orbits are nearly the same.

Find: The possible order of principal energy states.

For a hydrogen like atom, the speed of electron in the $n^{\text{th}}$ orbit is given by

v_n \propto \frac{1}{n}

So, writing the ratio of speeds,

\frac{v_1}{v_2}=\frac{3\times10^{5}}{2.5\times10^{5}}=\frac{6}{5}

Since $v_n\propto \frac{1}{n}$ ,

\frac{v_1}{v_2}=\frac{n_2}{n_1}

Therefore,

\frac{n_2}{n_1}=\frac{6}{5}

Thus, the possible values of $n_1$ and $n_2$ are $5$ and $6$ in the same ratio, so the order of energy states corresponding to the given electrons is $6$ and $5$ respectively.

The correct option is C.

Common mistakes

Using $v_n\propto n$ instead of $v_n\propto \frac{1}{n}$ . This reverses the relation between speed and orbit number. Always recall from the Bohr model that higher $n$ means lower electron speed.
Equating $\frac{v_1}{v_2}$ directly to $\frac{n_1}{n_2}$ . This is wrong because speed is inversely proportional to principal quantum number. Use $\frac{v_1}{v_2}=\frac{n_2}{n_1}$ instead.
Ignoring the word 'respectively' while matching the two electrons with their states. After obtaining the ratio, assign the first state to the first speed and the second state to the second speed carefully.

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