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).

The expression of energy is given by,

E_n = - [2.18× 10^-18] Z²/n²

Where,

Z = atomic number of the atom

N = principal quantum number

For ionization from n1 = 5 to n₂ = ∞,

Therefore ∆E = E2-E1 = -21.8× 10^-19× [1/n₂²-1/n₁²]

= 21.8× 10^-19 × [1/n₂²-1/n₁²]

=21.8× 10^-19× [1/5²-1/∞]

=8.72× 10^-20J

For ionization from 1^st orbit, n₁=1, n₂=∞

Therefore ∆E’ = 21.8× 10^-19× [1/1²- 1/∞]

= 21.8× 10^-19J

Now ∆E’/∆E = 21.8× 10^-19/8.72× 10^-20 = 25

Thus, the energy required to remove electron from 1^st orbit is 25 times than the required to electron from 5^th orbit.