Question 14 of 62intermediate🔧 ApplyNumerical3 marks

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).

Correct Answer

The energy required to ionise a H atom when the electron occupies the n=5n = 5 orbit is 8.72×10−208.72 \times 10^{-20} J. The ionization enthalpy of a H atom, which is the energy required to remove the electron from the n=1n = 1 orbit, is 2.18×10−182.18 \times 10^{-18} J. Comparing the two values, the energy required to ionise the atom from the n=5n = 5 orbit is exactly 125\frac{1}{25}th of the ionization enthalpy required from the ground state (n=1n = 1). This shows that the electron is less tightly bound in higher orbits.

Exercise: EXERCISES | Q: 2.14 | (Page 41)
For More Understanding

Explanation

Ionization energy is the energy required to move an electron from its current orbit to infinity (n=∞n = \infty), where the energy is zero. Using the formula En=−2.18×10−18(1n2)JE_n = -2.18 \times 10^{-18} \left( \frac{1}{n^2} \right) J, we calculate the energy for n=5n=5 and n=1n=1. The difference between E∞E_{\infty} (which is 0) and EnE_n gives the ionization energy for that orbit.

Solution Steps

  1. Step 1: Calculate energy of electron in n=5n = 5 orbit The energy of an electron in a hydrogen atom is given by En=−2.18×10−18(1n2)JE_n = -2.18 \times 10^{-18} \left( \frac{1}{n^2} \right) J. For n=5n = 5, E5=−2.18×10−18(152)J=−8.72×10−20E_5 = -2.18 \times 10^{-18} \left( \frac{1}{5^2} \right) J = -8.72 \times 10^{-20} J.

  2. Step 2: Calculate ionization energy from n=5n = 5 Ionization means moving the electron to n=∞n = \infty, where E∞=0E_{\infty} = 0. Energy required = E∞−E5=0−(−8.72×10−20)=8.72×10−20E_{\infty} - E_5 = 0 - (-8.72 \times 10^{-20}) = 8.72 \times 10^{-20} J.

  3. Step 3: Calculate ionization enthalpy from n=1n = 1 For n=1n = 1, E1=−2.18×10−18(112)J=−2.18×10−18E_1 = -2.18 \times 10^{-18} \left( \frac{1}{1^2} \right) J = -2.18 \times 10^{-18} J. Ionization enthalpy = E∞−E1=0−(−2.18×10−18)=2.18×10−18E_{\infty} - E_1 = 0 - (-2.18 \times 10^{-18}) = 2.18 \times 10^{-18} J.

  4. Step 4: Compare both values The energy required to ionise from n=5n = 5 is 8.72×10−202.18×10−18=125\frac{8.72 \times 10^{-20}}{2.18 \times 10^{-18}} = \frac{1}{25} of the ionization enthalpy from n=1n = 1.

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