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Atoms and Nuclei Part 3

Section: Physics

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Question : 16 of 98

Marks: +1, -0

Two electrons are moving in orbits of two hydrogen like atoms with speeds

3 \times 10^5 \,\text{m/s}

2.5 \times 10^5 \,\text{m/s}

respectively. If the radii of these orbits are nearly same then the possible order of energy states are

\_\_\_\_

[24 jan 2026 Shift 1]

8 and 10
10 and 12
9 and 8
6 and 5

Solution:

Concept:

For hydrogen-like atoms, the speed of an electron in an orbit is

v \propto \frac{Z}{n}

, and the radius is

r \propto \frac{n^2}{Z}

.

Explanation:

Given

v_1 = 3 \times 10^5 \,\text{m/s}

and

v_2 = 2.5 \times 10^5 \,\text{m/s}

.

Radii are equal:

r_1 = r_2

.

From

r \propto \frac{n^2}{Z}

, we get

\frac{n_1^2}{Z_1} = \frac{n_2^2}{Z_2}

, so

\frac{Z_1}{Z_2} = \left(\frac{n_1}{n_2}\right)^2

.

From

v \propto \frac{Z}{n}

, we get

\frac{v_1}{v_2} = \frac{Z_1}{n_1} \cdot \frac{n_2}{Z_2}

.

Substitute

\frac{Z_1}{Z_2}

:

\frac{v_1}{v_2} = \left(\frac{n_1}{n_2}\right)^2 \cdot \frac{n_2}{n_1} = \frac{n_1}{n_2}

.

Thus

\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{6}{5}

.

The simplest integer ratio is

n_1 : n_2 = 6 : 5

.

Answer:

6 and 5. (Option D)

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