Explanation:
Energy necessary to remove an electron from outer orbital of a neutral gaseous atom or molecule is known as ionization energy.
Bohr's equation to calculate energy is as follows.
[tex]\Delta E = R_{H}[\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}][/tex]
where, [tex]R_{H}[/tex] = Reydberg's constant = [tex]1.09 \times 10^{7}[/tex] per meter
In the given case, [tex]n_{1}[/tex] = 1 and [tex]n_{2}[/tex] = [tex]\infty[/tex]
Therefore, calculate the energy as follows.
[tex]\Delta E = R_{H}[\frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}}][/tex]
= [tex]1.09 \times 10^{7}[\frac{1}{(1)^{2}} - \frac{1}{\infty}}][/tex] J
= [tex]1.09 \times 10^{7}[1 - 0}][/tex] J
= [tex]1.09 \times 10^{7}][/tex] J
As atomic configuration of helium is [tex]1s^{2}[/tex]. Thus, energy required to produce [tex]He^{+}[/tex] and [tex]He^{2+}[/tex] will be the same.
Also, 1 joule = 0.239 calories. Hence, convert calculated energy into joules as follows.
[tex]1.09 \times 0.239 \times 10^{7}[/tex] J
= [tex]0.26051 \times 10^{7}[/tex] cal
= [tex]26.051 \times 10^{5}[/tex] cal
Therefore, we can conclude that [tex]26.051 \times 10^{5}[/tex] cal energy is required to produce, from neutral He atoms,1 mole of [tex]He^{+}[/tex] ions and [tex]He^{2+}[/tex] ions using Bohr's equations.
