Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \mathrm{~J}$. Calculate the ionization enthalpy of atomic hydrogen in terms of $\mathrm{J}$ mol $^{-1}$.
It is given that the energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18} \mathrm{~J}$.
Therefore, the energy required to remove that electron from the ground state of hydrogen atom is $2.18 \times 10^{-18} \mathrm{~J}$.
$\therefore$ Ionization enthalpy of atomic hydrogen $=2.18 \times 10^{-18} \mathrm{~J}$
Hence, ionization enthalpy of atomic hydrogen in terms of $\mathrm{J} \mathrm{mol}^{-1}=2.18 \times 10^{-18} \times 6.02 \times 10^{23} \mathrm{~J} \mathrm{~mol}^{-1}=1.31 \times 10^{6} \mathrm{~J} \mathrm{~mol}^{-1}$
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