Respuesta :
Since you have the ionization energy for 1 mole of hydrogen atoms, just divide that by Avogadro's number to get energy/atom:
1.31 X 10^6 J/mol / 6.02 X 10^23 atoms/mol = 2.18 X 10^-18 J/atom
1.31 X 10^6 J/mol / 6.02 X 10^23 atoms/mol = 2.18 X 10^-18 J/atom
The ionization energy for [tex]{\text{H}}{{\text{e}}^ + }[/tex] is [tex]\boxed{1.0974 \times {{10}^7}{\text{ J}}}[/tex].
Further explanation:
Ionization energy:
It is the amount of energy needed for removal of the most loosely bound valence electrons from the isolated neutral gaseous atom. It is usually denoted by IE. It depends on the ease of electron removal from neutral atoms. More is the ease of electron removal, less is the ionization energy and vice-versa.
The expression for Rydberg equation is as follows:
[tex]\Delta E = {R_{\text{H}}}\left( {\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)[/tex] …… (1)
Where,
[tex]\Delta E[/tex] is the energy difference between two energy levels.
[tex]{R_{\text{H}}}[/tex] is Rydberg constant.
[tex]{{\text{n}}_{\text{1}}}[/tex] is the lower energy level.
[tex]{{\text{n}}_{\text{2}}}[/tex] is the higher energy level.
Since [tex]{\text{H}}{{\text{e}}^ + }[/tex] is formed by removal of one electron from neutral He atom, it has left with only one shell and therefore principal quantum number becomes 1.
Substitute 1 for [tex]{{\text{n}}_{\text{1}}}[/tex] , [tex]1.0974 \times {10^7}{\text{ }}{{\text{m}}^{ - 1}}[/tex] for [tex]{R_{\text{H}}}[/tex] and [tex]\infty[/tex] for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (1) to calculate ionization energy for [tex]{\text{H}}{{\text{e}}^ + }[/tex] ion.
[tex]\begin{aligned}\Delta E &= \left( {1.0974 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( {\text{1}} \right)}^2}}} - \frac{1}{{{{\left( \infty \right)}^2}}}} \right)\\&= 1.0974 \times {10^7}{\text{ J}}\\\end{aligned}[/tex]
Therefore the value of ionization energy for [tex]{\text{H}}{{\text{e}}^ + }[/tex] ion is [tex]1.0974 \times {10^7}{\text{ J}}[/tex].
Learn more:
- Rank the elements according to first ionization energy: https://brainly.com/question/1550767
- Write the chemical equation for the first ionization energy of lithium: https://brainly.com/question/5880605
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Periodic classification of elements
Keywords: ionization energy, IE, energy, most loosely, isolated, neutral, atom, ionization energy, He+, n1, n2, energy difference, principal quantum number.
