Question

In a hydrogen-like ion, the energy difference between the 2nd excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

• A. 4
• B. 2
• C. 1
• D. 3

Hint:

In hydrogen-like ions, the energy levels follow the formula \( E_n = - \frac{13.6 Z^2}{n^2} \). Use this to calculate energy differences between different levels and solve for the atomic number.

Answer 1

The Correct Option is D

The energy levels in a hydrogen-like ion are described by \( E_n = - \frac{13.6 \, \text{eV} \times Z^2}{n^2} \), where \( E_n \) denotes the energy of the \( n^{th} \) level, \( Z \) is the atomic number, and \( n \) is the principal quantum number (taking values 1, 2, 3, etc.). The energy difference between the 2nd excitation state (\( n = 3 \)) and the ground state (\( n = 1 \)) is provided as 108.8 eV. The energy for the \( n = 3 \) state is \( E_3 = - \frac{13.6 Z^2}{9} \), and the energy for the \( n = 1 \) state (ground state) is \( E_1 = - 13.6 Z^2 \). The energy difference \( \Delta E \) is calculated as \( \Delta E = E_1 - E_3 = - 13.6 Z^2 - \left( - \frac{13.6 Z^2}{9} \right) = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) = \frac{108.8 Z^2}{9} \). Given \( \Delta E = 108.8 \, \text{eV} \), we set up the equation \( \frac{108.8 Z^2}{9} = 108.8 \), which simplifies to \( Z^2 = 9 \). Therefore, the atomic number \( Z \) is 3.