Chapter 27, Problem 51P

To determine

The ionization energy of doubly ionized lithium atom.

Answer to Problem 51P

Solution:

The ionization energy of doubly ionized lithium atom is $122\text{eV}$.

Explanation of Solution

Doubly ionized lithium atom $(L{i}^{+2})$ having 1 electron and 3 protons in the nucleus of the lithium atom. The square of product of charges appears in the energy levels

Energy required for the ionization can be calculated by using the formula $E=\frac{z^2(13.6)}{n^2}$

$E=\frac{z^2(13.6)}{n^2}$, where z = number of protons (3)and n= number of electron (1)

So, $E=\frac{z^2(13.6)}{n^2}=\frac{3^2(13.6)}{1^2}=122\text{eV}$

Energy required for the ionization is $122\text{eV}$.