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,
\(E_n = \frac {-(2.18×10^{-18}) Z^2}{n^2}\)
Where,
Z = atomic number of the atom
n= principal quantum number
For ionization from n₁= 5 to n₂ = ∞
\(ΔE = E_∞ - E_5\)
\(ΔE= [\frac {-(2.18×10^{-18} J) (1)^2}{(∞)^2}] - [{ \frac {-(2.18×10^{-18} J) (1)^2}{(5)^2}}]\)
\(ΔE= -(2.18×10^{-18} J) (\frac {1}{5^2 })\)           (Since \(\frac {1}{∞} = 0\))
\(ΔE= 0.0872×10^{-18} J\)
\(ΔE = 0.0872×10^{-20} J\)
Hence, the energy required for ionization from n=5 to n=∞ is \(8.72 × 10^{-20} J\).
Energy required for n₁=1 to n=∞ ,
\(ΔE = E_∞ - E_1\)
\(ΔE= [\frac {-(2.18×10^{-18} J) (1)^2}{(∞)^2}] - [{ \frac {-(2.18×10^{-18} J) (1)^2}{(1)^2}}]\)
\(ΔE = (2.18 × 10^{-18}) [0+1]\)
\(ΔE = (2.18 × 10^{-18}) J\)

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