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Write Brief Answer · Q8

Q.Energy of an electron in the ground state of the hydrogen atom is -2.18 × 10⁻¹⁸ J. Calculate the ionisation enthalpy of atomic hydrogen in terms of kJ mol⁻¹.

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Step 1. The ground-state electron's energy is E=−2.18×10−18E = -2.18 \times 10^{-18} J (negative, since it is bound to the nucleus). Removing it completely (to a state of zero energy, i.e. E = 0, infinitely far from the nucleus) requires supplying the negative of this energy:

IEper atom=0−(−2.18×10−18)=2.18×10−18 J per atomIE_{\text{per atom}} = 0 - (-2.18\times10^{-18}) = 2.18\times10^{-18}\ \text{J per atom}

Step 2. Convert to a per-mole quantity by multiplying by Avogadro's number, NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\ \text{mol}^{-1}: …

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