Question

The ratio of ground state energy of \( \text{Li}^{2+}, \text{He}^{+}, \text{H} \) is:

• A. \( 3:2:1 \)
• B. \( 1:2:3 \)
• C. \( 9:4:1 \)
• D. \( 1:4:9 \)

Hint:

The ground state energy of a hydrogen-like ion is proportional to \( Z^2 \). To determine ratios, simply square the atomic number of the respective species.

Answer 1

The Correct Option is C

Step 1: Understanding Ground State Energy Formula 
The ground state energy (\(E_n\)) of a hydrogen-like ion is given by the formula: \[ E_n = - \frac{13.6 Z^2}{n^2} \text{ eV} \] where: - \( Z \) is the atomic number, - \( n \) is the principal quantum number (for ground state, \( n = 1 \)). 

Step 2: Calculating for Each Ion 
- For Hydrogen (\( H \)), \( Z = 1 \): \[ E_H = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV} \] - For Helium ion (\( He^+ \)), \( Z = 2 \): \[ E_{He^+} = -13.6 \times \frac{2^2}{1^2} = -54.4 \text{ eV} \] - For Lithium ion (\( Li^{2+} \)), \( Z = 3 \): \[ E_{Li^{2+}} = -13.6 \times \frac{3^2}{1^2} = -122.4 \text{ eV} \] 

Step 3: Finding the Ratio 
\[ E_{Li^{2+}} : E_{He^+} : E_H = 9:4:1 \]