Question

The difference between the radii of \(3^{rd}\) and \(2^{nd}\) orbit of H-atom is x pm. The difference between the radii of \(4^{th}\) and \(3^{rd}\) orbit of \( \text{Li}^{2+} \) ion is y pm. \(y:x\) is equal to

• A. 15:7
• B. 7:15
• C. 3:1
• D. 1:3

Hint:

Bohr radius for n-th orbit of a hydrogen-like atom (atomic number Z): \( r_n = a_0 \frac{n^2}{Z} \), where \(a_0\) is the Bohr radius (\( \approx 0.529 \, \text{\AA} = 52.9 \, \text{pm} \)). Calculate the radii for the specified orbits and atoms/ions. Find the differences \(x\) and \(y\). Calculate the ratio \(y:x\). For H-atom, Z=1. For \( \text{Li}^{2+} \), Z=3.

Answer 1

The Correct Option is B

The radius of the n-th orbit in a hydrogen-like atom (atomic number Z) is given by Bohr's model: \[ r_n = a_0 \frac{n^2}{Z} \] where \( a_0 \) is the Bohr radius (approximately 52.
9 pm).
For H-atom, \( Z=1 \).
Radius of \(3^{rd}\) orbit: \( r_{H,3} = a_0 \frac{3^2}{1} = 9a_0 \).
Radius of \(2^{nd}\) orbit: \( r_{H,2} = a_0 \frac{2^2}{1} = 4a_0 \).
Difference \( x = r_{H,3} - r_{H,2} = 9a_0 - 4a_0 = 5a_0 \).
For \( \text{Li}^{2+} \) ion, Lithium has atomic number \( Z=3 \).
Radius of \(4^{th}\) orbit: \( r_{Li,4} = a_0 \frac{4^2}{3} = \frac{16}{3}a_0 \).
Radius of \(3^{rd}\) orbit: \( r_{Li,3} = a_0 \frac{3^2}{3} = \frac{9}{3}a_0 = 3a_0 \).
Difference \( y = r_{Li,4} - r_{Li,3} = \frac{16}{3}a_0 - 3a_0 = \left(\frac{16}{3} - \frac{9}{3}\right)a_0 = \frac{16-9}{3}a_0 = \frac{7}{3}a_0 \).
We need the ratio \( y:x \).
\[ \frac{y}{x} = \frac{\frac{7}{3}a_0}{5a_0} = \frac{7/3}{5} = \frac{7}{3 \times 5} = \frac{7}{15} \] So, \( y:x = 7:15 \).
This matches option (2).