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Chemistry Question on Quantum Mechanical Model of Atom

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

The expression of energy is given by,
En=−(2.18×10−18)Z2n2E_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∞−E5ΔE = E_∞ - E_5
ΔE=[−(2.18×10−18J)(1)2(∞)2]−[−(2.18×10−18J)(1)2(5)2]ΔE= [\frac {-(2.18×10^{-18} J) (1)^2}{(∞)^2}] - [{ \frac {-(2.18×10^{-18} J) (1)^2}{(5)^2}}]
ΔE=−(2.18×10−18J)(152)ΔE= -(2.18×10^{-18} J) (\frac {1}{5^2 }) (Since 1∞=0\frac {1}{∞} = 0)
ΔE=0.0872×10−18JΔE= 0.0872×10^{-18} J
ΔE=0.0872×10−20JΔE = 0.0872×10^{-20} J
Hence, the energy required for ionization from n=5 to n=∞ is 8.72×10−20J8.72 × 10^{-20} J.
Energy required for n1=1 to n=∞ ,
ΔE=E∞−E1ΔE = E_∞ - E_1
ΔE=[−(2.18×10−18J)(1)2(∞)2]−[−(2.18×10−18J)(1)2(1)2]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}) [0+1]
ΔE=(2.18×10−18)JΔE = (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.