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

In a hydrogen-like ion, the energy levels are given by the formula: \[ E_n = - \frac{13.6 \, \text{eV} \times Z^2}{n^2} \] where: \( E_n \) is the energy of the \( n^{th} \) level, \( Z \) is the atomic number, \( n \) is the principal quantum number (1, 2, 3, etc.). 
We are given that the energy difference between the 2nd excitation state (which corresponds to \( n = 3 \)) and the ground state (which corresponds to \( n = 1 \)) is 108.8 eV. The energy for the \( n = 3 \) state is: \[ E_3 = - \frac{13.6 \times Z^2}{3^2} = - \frac{13.6 Z^2}{9} \] The energy for the \( n = 1 \) state (ground state) is: \[ E_1 = - \frac{13.6 Z^2}{1^2} = - 13.6 Z^2 \] The energy difference \( \Delta E \) between the 2nd excitation state and the ground state is: \[ \Delta E = E_1 - E_3 = - 13.6 Z^2 - \left( - \frac{13.6 Z^2}{9} \right) \] \[ \Delta E = - 13.6 Z^2 + \frac{13.6 Z^2}{9} = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) \] \[ \Delta E = 13.6 Z^2 \times \frac{8}{9} = \frac{108.8 Z^2}{9} \] We are given that \( \Delta E = 108.8 \, \text{eV} \), so: \[ \frac{108.8 Z^2}{9} = 108.8 \] \[ Z^2 = 9 \quad \Rightarrow \quad Z = 3 \] 
Thus, the atomic number of the ion is 3.