The ionization enthalpy of `He^(+)` ion is `19.60 xx 10^(-18) J "atom"^(-1)`. The ionization enthalpy of `Li^(2+)` ion will be

Options

A. `84.2 xx 10^(-18) J "atom"^(-1)`
B. `44.10 xx 10^(-18) J "atom"^(-1)`
C. `63.20 xx 10^(-18) J "atom"^(-1)`
D. `21.20 xx 10^(-18) J atom^(-1)`

Explanation

To find the ionization enthalpy of the `Li^(2+)` ion, we can use the concept of energy levels in hydrogen-like atoms as described by Bohr's model. The ionization enthalpy can be derived from the formula for the energy of an electron in a hydrogen-like atom:

### Step-by-Step Solution:

1. **Understand the Formula**:
The energy of an electron in the nth orbit of a hydrogen-like atom is given by:
\[ E_n = -\frac{2 \pi^2 m e^4 Z^2}{n^2 h^2} \]
where \( Z \) is the atomic number, \( n \) is the principal quantum number, \( m \) is the mass of the electron, \( e \) is the charge of the electron, and \( h \) is Planck's constant.

2. **Identify Constants**:
For hydrogen-like species, we can simplify the expression by defining a constant \( k \):
\[ k = \frac{2 \pi^2 m e^4}{h^2} \]
Thus, the energy can be rewritten as:
\[ E_n = -k \frac{Z^2}{n^2} \]

3. **Ionization Energy Calculation for `He^(+)`**:
For the `He^(+)` ion (where \( Z = 2 \)):
- The ionization enthalpy is given as \( 19.60 \times 10^{-18} \, \text{J/atom} \).
- The ionization energy can be expressed as:
\[ E_{\text{ionization}} = E_{\infty} - E_1 = 0 - (-k \frac{Z^2}{1^2}) = k Z^2 \]
Thus, for `He^(+)`:
\[ E_{\text{ionization}} = 4k \]
Therefore, we can find \( k \):
\[ 4k = 19.60 \times 10^{-18} \implies k = \frac{19.60 \times 10^{-18}}{4} = 4.90 \times 10^{-18} \, \text{J/atom} \]

4. **Ionization Energy Calculation for `Li^(2+)`**:
For the `Li^(2+)` ion (where \( Z = 3 \)):
- The ionization energy can be calculated as:
\[ E_{\text{ionization}} = k Z^2 = k \cdot 3^2 = 9k \]
Substituting the value of \( k \):
\[ E_{\text{ionization}} = 9 \cdot 4.90 \times 10^{-18} = 44.10 \times 10^{-18} \, \text{J/atom} \]

5. **Final Result**:
The ionization enthalpy of `Li^(2+)` is:
\[ \boxed{44.10 \times 10^{-18} \, \text{J/atom}} \]