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erastova [34]
3 years ago
5

The ionization energy for a hydrogen atom is 1.31×106 J/mol. What is the ionization energy for He+?

Chemistry
2 answers:
Sidana [21]3 years ago
7 0

The ionization energy for {\text{H}}{{\text{e}}^ + } is \boxed{1.0974 \times {{10}^7}{\text{ J}}}.

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:

\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)                                                        …… (1)

Where,

\Delta E is the energy difference between two energy levels.

{R_{\text{H}}} is Rydberg constant.

{{\text{n}}_{\text{1}}} is the lower energy level.

{{\text{n}}_{\text{2}}} is the higher energy level.

Since {\text{H}}{{\text{e}}^ + } 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 {{\text{n}}_{\text{1}}} , 1.0974 \times {10^7}{\text{ }}{{\text{m}}^{ - 1}} for {R_{\text{H}}} and \infty for {{\text{n}}_{\text{2}}}  in equation (1) to calculate ionization energy for {\text{H}}{{\text{e}}^ + } ion.

 \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}

Therefore the value of ionization energy for {\text{H}}{{\text{e}}^ + } ion is 1.0974 \times {10^7}{\text{ J}}.

Learn more:

  1. Rank the elements according to first ionization energy: brainly.com/question/1550767
  2. Write the chemical equation for the first ionization energy of lithium: 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.

miss Akunina [59]3 years ago
3 0
Since you have the ionization energy for 1 mole of hydrogen atoms, just divide that by Avogadro's number to get energy/atom: 

<span>1.31 X 10^6 J/mol / 6.02 X 10^23 atoms/mol = 2.18 X 10^-18 J/atom</span>
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Answer:

1.58x10⁻⁵

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Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

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log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

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iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

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[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

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