Question

The screened nuclear charge of neutral Helium atom is given as \( 1.7e \), where \( e \) is the magnitude of the electronic charge. Assuming the Bohr model of the atom for which the energy levels are \( E_n = - \frac{Z^2}{2n^2} \) atomic units (where \( Z \) is the atomic number), the first ionization potential of Helium in atomic units is:

• A. 0.89
• B. 1.78
• C. 0.94
• D. 3.16

Hint:

When dealing with ionization potentials in the Bohr model, remember that the energy levels of electrons are quantized and depend on the effective nuclear charge \( Z_{{eff}} \).

Answer 1

The Correct Option is A

Step 1: The effective nuclear charge is \( Z_{\text{eff}} = 1.7 \), so the atomic number \( Z \) can be assumed to be 2.
\[ E_n = - \frac{Z^2}{2n^2}. \] For the first ionization, the electron is removed from the \( n = 1 \) energy level, so the energy of the first ionization is: \[ E_1 = - \frac{Z_{\text{eff}}^2}{2(1)^2} = - \frac{(1.7)^2}{2} = - 1.445 \, \text{atomic units}. \] The ionization potential is the absolute value of the energy, so the first ionization potential is: \[ \text{Ionization potential} = 1.445 \, \text{atomic units}. \] Step 2: The ionization potential is approximately \( 0.89 \) atomic units.