Question

Which of the following have same radius according to Bohr's theory : (A) Radius of \(1^{st}\) orbit of H-atom.
(B) Radius of \(1^{st}\) orbit of He\(^{+}\).
(C) Radius of \(II^{nd}\) orbit of He\(^{2+}\).
(D) Radius of \(II^{nd}\) orbit of Li\(^{2+}\).
(E) Radius of \(II^{nd}\) orbit of Be\(^{3+}\).

• A. A and B
• B. A and E
• C. B, C and E
• D. A and D

Answer 1

The Correct Option is B

To determine which of the provided options have the same radius according to Bohr's theory, let's analyze the radii of the given orbits using the Bohr's formula for the radius of an electron orbit:

\(r_n = \frac{n^2 \cdot h^2}{4 \pi^2 \cdot m \cdot e^2 \cdot Z}\)

Where:

• \(r_n\) is the radius of the orbit.
• \(n\) is the principal quantum number of the orbit.
• \(h\) is Planck's constant.
• \(m\) is the mass of the electron.
• \(e\) is the charge of the electron.
• \(Z\) is the atomic number of the element.

The radius of the nth orbit in a hydrogen-like (single-electron) ion is also given by:

\(r_n = \frac{n^2 \cdot a_0}{Z}\)

Where \(a_0\) is the Bohr radius (approximately 0.529 Å).

1. For H-atom in the \(1^{st}\) orbit (A): \(n = 1\), \(Z = 1\)

 

\(r_1 = \frac{1^2 \cdot a_0}{1} = a_0\)

1. For He\(^{+}\) in the \(1^{st}\) orbit (B): \(n = 1\), \(Z = 2\)

 

\(r_1 = \frac{1^2 \cdot a_0}{2} = \frac{a_0}{2}\)

1. For He\(^{2+}\) in the \(II^{nd}\) orbit (C): \(n = 2\), \(Z = 2\)

 

\(r_2 = \frac{2^2 \cdot a_0}{2} = 2a_0\)

1. For Li\(^{2+}\) in the \(II^{nd}\) orbit (D): \(n = 2\), \(Z = 3\)

 

\(r_2 = \frac{2^2 \cdot a_0}{3} = \frac{4a_0}{3}\)

1. For Be\(^{3+}\) in the \(II^{nd}\) orbit (E): \(n = 2\), \(Z = 4\)

 

\(r_2 = \frac{2^2 \cdot a_0}{4} = a_0\)

By comparing these radii, we find that:

• The radius of the \(1^{st}\) orbit of H-atom (A) is \(a_0\).
• The radius of the \(II^{nd}\) orbit of Be\(^{3+}\) (E) is \(a_0\).

Thus, options A and E have the same radius according to Bohr's theory. Therefore, the correct answer is: A and E.