How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?

Solution:

The expression of energy is given by,

$E_{n}=\frac{-\left(2.18 \times 10^{-18}\right) Z^{2}}{n^{2}}$

Where,

Z = atomic number of the atom n

= principal quantum number

For ionization from $n_{1}=5$ to $n_{2}=\infty$,

$\Delta E=E_{x}-E_{3}$

$=\left[\left\{\frac{-\left(2.18 \times 10^{-18} \mathrm{~J}\right)(1)^{2}}{(\infty)^{2}}\right\}-\left\{\frac{-\left(2.18 \times 10^{-18} \mathrm{~J}\right)(1)^{2}}{(5)^{2}}\right\}\right]$

$=\left(2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{1}{(5)^{2}}\right) \quad\left(\right.$ Since $\left.\frac{1}{\infty}=0\right)$

$=0.0872 \times 10^{-18} \mathrm{~J}$

$\Delta E=8.72 \times 10^{-20} \mathrm{~J}$

Hence, the energy required for ionization from $n=5$ to $n=\infty$

Energy required for $n_{1}=1$ to $n=\infty$,

is $8.72 \times 10^{-20}$ ].

$\Delta E^{\prime}=E_{\infty}-E_{1}$

$=\left[\left\{\frac{-\left(2.18 \times 10^{-18}\right)(1)^{2}}{(\infty)^{2}}\right\}-\left\{\frac{-\left(2.18 \times 10^{-18}\right)(1)^{2}}{(1)^{2}}\right\}\right]$

$=\left(2.18 \times 10^{-18}\right)[1-0]$

$=2.18 \times 10^{-18} \mathrm{~J}$

Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.