Question

Energy of an electron in the ground state of the hydrogen atom is $ - 2.18 \times {10^{ - 18}}J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $Jmo{l^{ - 1}}$.

Answer 1

Hint: Ionization energy is the energy required to remove an electron from ground state to infinity. Thus, ionization enthalpy can be calculated as the difference between the energy of an electron in its ground state and the energy of an electron at infinity.

Complete step by step answer:
The ionization enthalpy can be defined as the amount of energy required to remove an electron from an isolated gaseous atom in its ground state to infinity.
Thus, it can be calculated as the difference between the energy of an electron in its ground state (say,${E_1}$) and the energy of an electron at infinity (${E_\infty }$). Mathematically,
${E_\infty } - {E_1}$
We are given that the energy of an electron in the ground state of the hydrogen atom is $ - 2.18 \times {10^{ - 18}}J$.
Thus, ${E_1}$ for hydrogen atom = $ - 2.18 \times {10^{ - 18}}J$.
Now, the energy of an electron at infinity is zero. This means, ${E_\infty }$ for hydrogen atom = 0.
Therefore, ionization enthalpy of atomic hydrogen (in joules):
${E_\infty } - {E_1} = 0 - ( - 2.18 \times {10^{ - 18}}J) = 2.18 \times {10^{ - 18}}J$.

Therefore, the energy required to remove an electron from a hydrogen atom in terms of joule is $2.18 \times {10^{ - 18}}J$. But we are asked the ionization enthalpy of atomic hydrogen in terms of $Jmo{l^{ - 1}}$. This means we need to find the energy required to remove 1 mole of electrons.
Since, 1 mole of any substance contains Avogadro's number of particles i.e., $6.022 \times {10^{23}}$ particles.
Therefore, the amount of energy required to remove 1 mole of electrons =
$2.18 \times {10^{ - 18}} \times 6.022 \times {10^{23}} = 13.30 \times {10^5}{\text{ }}Jmo{l^{ - 1}}$.
Hence, this is the required ionization enthalpy of hydrogen atom.

Note: Quantitatively, ionization enthalpy for an element, let X can be expressed as:
$X \to {X^ + } + {e^ - }$
The SI unit of ionization energy is ${\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$. Take care of the units while doing the calculations of this question.