Question:

Bohr's radius of 2nd orbit of $ \text{Be}^{3+} $ is equal to that of

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The Bohr radius depends on the atomic number $ Z $ and the quantum number $ n $. For the same quantum number $ n $, a larger $ Z $ leads to a smaller orbit.

Updated On: May 3, 2025

4th orbit of hydrogen
2nd orbit of He$^{+} $
3rd orbit of Li$^{2+} $
1st orbit of hydrogen

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The Correct Option is D

Approach Solution - 1

To determine the equivalent Bohr's radius of the second orbit of $ \text{Be}^{3+} $, we employ the formula for Bohr's radius $ r_n $ of an orbit in a hydrogen-like atom, given by:

$ r_n = \frac{n^2 h^2}{4\pi^2 m e^2 Z} $, where $ n $ is the orbit number, $ h $ is Planck's constant, $ m $ is the electron mass, $ e $ is the electron charge, and $ Z $ is the atomic number of the element.

For a hydrogen-like ion, the formula further simplifies when compared to known Bohr's radius for hydrogen ($ r_n^{\text{H}} $):

$ r_n = \frac{n^2}{Z} \times r_n^{\text{H}} $.

For $ \text{Be}^{3+} $, the atomic number $ Z $ is 4. So for the second orbit ($ n = 2 $):

$ r_2 = \frac{2^2}{4} \times r_1^{\text{H}} = \frac{4}{4} \times r_1^{\text{H}} = r_1^{\text{H}} $.

This shows that the second orbit of $ \text{Be}^{3+} $ has a radius equal to the first orbit of hydrogen.

This precisely matches the option: 1st orbit of hydrogen.

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Approach Solution -2

The radius of an electron's orbit in Bohr's model is given by the formula: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z} \] Where:
- $ n $ is the principal quantum number,
- $ h $ is Planck's constant,
- $ \epsilon_0 $ is the permittivity of free space,
- $ m $ is the mass of the electron,
- $ e $ is the charge of the electron, and
- $ Z $ is the atomic number of the ion. We are asked to compare the Bohr radius of the 2nd orbit of $ \text{Be}^{3+} $ and various orbits of other atoms. For $ \text{Be}^{3+} $, the atomic number is $ Z = 4 $, and for the second orbit, $ n = 2 $. For hydrogen, $ Z = 1 $, and for the first orbit, $ n = 1 $.

The formula shows that the radius of the orbit is inversely proportional to $ Z $, and directly proportional to $ n^2 $.
Thus, the Bohr radius for the second orbit of $ \text{Be}^{3+} $ can be compared with the first orbit of hydrogen, as they both satisfy the relationship of proportionality when adjusted for their respective values of $ n $ and $ Z $.

Therefore, the Bohr radius of the 2nd orbit of $ \text{Be}^{3+} $ is equal to the Bohr radius of the 1st orbit of hydrogen.
Thus, the correct answer is 1st orbit of hydrogen.