Question

Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18  J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1  .

Answer 1

Given

Energy of electron in ground state of H‑atom: \[ E_1 = -2.18 \times 10^{-18} \ \text{J per atom} \] Ionization means taking the electron from \(n = 1\) to \(n = \infty\), where \[ E_\infty = 0 \ \text{J} \]

Step 1: Energy Required per Atom

\[ \Delta E_{\text{per atom}} = E_\infty - E_1 = 0 - \left(-2.18 \times 10^{-18}\right) = 2.18 \times 10^{-18} \ \text{J} \]

Step 2: Convert to per mole

Use Avogadro constant: \[ N_A = 6.022 \times 10^{23} \ \text{mol}^{-1} \] \[ \Delta H_{\text{ion}} = \Delta E_{\text{per atom}} \times N_A = \left(2.18 \times 10^{-18}\right)\left(6.022 \times 10^{23}\right) \ \text{J mol}^{-1} \]

Step 3: Numerical Value

Multiply: \[ 2.18 \times 6.022 \approx 13.1 \] and \[ 10^{-18} \times 10^{23} = 10^{5} \] So: \[ \Delta H_{\text{ion}} \approx 13.1 \times 10^{5} \ \text{J mol}^{-1} = 1.31 \times 10^{6} \ \text{J mol}^{-1} \]

Final Answer

The ionization enthalpy of atomic hydrogen is \[ \boxed{1.31 \times 10^{6} \ \text{J mol}^{-1}} \]