Question

Ionisation energy of \( \text{He}^+ \) is \( 19.6 \times 10^{-18} \, \text{J} \, \text{atom}^{-1} \). The energy of the first stationary state (n = 1) of \( \text{Li}^{2+} \) is:

• A. \( 4.41 \times 10^{-16} \, \text{J} \, \text{atom}^{-1} \)
• B. \( -4.41 \times 10^{-17} \, \text{J} \, \text{atom}^{-1} \)
• C. \( -2.2 \times 10^{-15} \, \text{J} \, \text{atom}^{-1} \)
• D. \( 8.82 \times 10^{-17} \, \text{J} \, \text{atom}^{-1} \)

Hint:

For hydrogen-like atoms, the energy of the \( n \)-th level is calculated using the formula \( E_n = -13.6 \, Z^2/n^2 \, \text{eV} \).

Answer 1

The Correct Option is A

Step 1: The ionisation energy for a hydrogen-like atom is given by: \[ E = -13.6 \, \dfrac{Z^2}{n^2} \, \text{eV}. \] For \( \text{Li}^{2+} \), \( Z = 3 \) and \( n = 1 \), so the energy is: \[ E = -13.6 \times \dfrac{3^2}{1^2} = -13.6 \times 9 = -122.4 \, \text{eV}. \] Step 2: Converting this to joules: \[ E = -122.4 \times 1.6 \times 10^{-19} = -1.958 \times 10^{-17} \, \text{J}. \]
Final Answer: \[ \boxed{4.41 \times 10^{-16} \, \text{J} \, \text{atom}^{-1}} \]