2. structure of atom — Chemistry STD 11 Science — Question

Mathematically,

$I \cdot E \cdot=E_{\infty}-E_{1}$

Energy associated to the $n^{t h}$ - energy level will be:

$E_{n}=\frac{E_{o} \times Z^{2}}{n^{2}}$

For ground state energy level of $L i^{2+}-i o n, n=1$ and $Z=3$

lonization energy $=122.4$ eV (given)

Putting the values in equation $1,$ we get

$122.4 e V=0-\frac{E_{o} \times 3^{2}}{1^{2}}$

$E_{o}=-13.6 \mathrm{eV}$

Now, we have to calculate the sixth ionization potential for carbon, we must know the principle quantum number for this atom.

Atomic number of Carbon $=6$

And we are asked to find the 6 th ionization potential which means that 5 electrons are already released and the electronic configuration for $C^{5+}$ is $1 s^{2}$

So, the principle quantum number for the 6 th ionization step $=1$

Putting the values in equation $1,$ we get $E_{1}=\frac{-13.6 \mathrm{eV} \times(6)^{2}}{(1)^{2}}=-489.6 \mathrm{eV}$

$E_{1}=-489.6 \mathrm{eV}$

Therefore, 6 th ionization potential for carbon atom is $-489.6 \mathrm{eV}$.