Question

How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Answer 1

The expression of energy is given by,
$E_0=\frac{-(2.18\times {10}^{-18})Z^2}{n^2}$
Where,
Z = atomic number of the atom
n = principal quantum number
For ionization from $n_1=5to{n}_2=\infty ,$
$△E=E_{\infty }-E_5=[\left\{\frac{-(2.18\times {10}^{-18}J)(1{)}^2}{(\infty {)}^2}\right\}-\left\{\frac{-(2.18\times {10}^{-18}J)(1{)}^2}{(5{)}^2}\right\}]=(2.18\times {10}^{-18}J)\left(\frac{1}{(5{)}^2}\right)(Since\frac{1}{\infty }=0)=0.0872\times {10}^{-18}J△E=8.72\times {10}^{-20}J$
Hence, the energy required for ionization from n = 5 to n = $\infty$ Energy required for n1 = 1 to n = $\infty$, is 8.72 × ${10}^{–20}$ J.
$△E=E_{\infty }-E_1=[\left\{\frac{-(2.18\times {10}^{-18})(1{)}^2}{\left(\infty \right)}\right\}-\left\{\frac{-(2.18\times {10}^{-18})(1{)}^2}{(1{)}^2}\right\}]=(2.18\times {10}^{-18})=2.18\times {10}^{-18}J$
Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.