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Chapter 6: Problem 147

Singly ionized atomic helium \(\mathrm{He}^{+1}\) is a hydrogen like ion. (a) What is its ground-state radius? (b) Calculate the energies of its four lowest energy states. (c) Repeat the calculations for the \(L i^{2+}\) ion.

Short Answer

Expert verified
The ground-state radius for He+1 ion is \(2.645886\times10^{-11}\:m\), and the energies of its four lowest energy states are -54.4 eV, -13.6 eV, -6.04 eV, and -3.4 eV. The ground-state radius for Li+2 ion is \(1.763924\times10^{-11}\:m\), and the energies of its four lowest energy states are -122.4 eV, -30.6 eV, -13.6 eV, and -7.65 eV.

Step by step solution

01

Identify the atomic number of He+1

He has an atomic number of 2 (\(Z = 2\)). The helium ion has lost an electron, becoming a singly-ionized atom, and its electron configuration is similar to a hydrogen atom.
02

Calculate the ground-state radius for He+1

Using the formula for the ground-state radius, \(r_1 = 1^2 \cdot \frac{a_0}{Z} = \frac{a_0}{2} = \frac{5.291772\times10^{-11}}{2} = 2.645886\times10^{-11}\:m\). So, the ground-state radius for He+1 ion is \(2.645886\times10^{-11}\:m\). ##(b) Energies of the four lowest energy states of He+1##
03

Calculate the energy levels for He+1

Using the formula for energy levels, we will determine the energy for the lowest \(n\) values: - For \(n=1\): \(E_1 = -\frac{2^2\,13.6}{1^2}=-54.4\text{ eV}\) - For \(n=2\): \(E_2 = -\frac{2^2\,13.6}{2^2}=-13.6\text{ eV}\) - For \(n=3\): \(E_3 = -\frac{2^2\,13.6}{3^2}=-6.04\text{ eV}\) - For \(n=4\): \(E_4 = -\frac{2^2\,13.6}{4^2}=-3.4\text{ eV}\) The energies of the four lowest energy states for He+1 are: -54.4 eV, -13.6 eV, -6.04 eV, and -3.4 eV. ##(c) Ground-state radius and energies of the four lowest energy states for Li+2##
04

Identify the atomic number of Li+2

Li has an atomic number of 3 (\(Z = 3\)). The lithium ion has lost two electrons, becoming a doubly-ionized atom, and its electron configuration is similar to a hydrogen atom.
05

Calculate the ground-state radius for Li+2

Using the formula for the ground-state radius, \(r_1 = 1^2\cdot \frac{a_0}{Z} = \frac{a_0}{3} = \frac{5.291772\times10^{-11}}{3} = 1.763924\times10^{-11}\:m\). So, the ground-state radius for Li+2 ion is \(1.763924\times10^{-11}\:m\).
06

Calculate the energy levels for Li+2

Using the formula for energy levels, we will determine the energy for the lowest \(n\) values: - For \(n=1\): \(E_1 = -\frac{3^2\,13.6}{1^2}=-122.4\text{ eV}\) - For \(n=2\): \(E_2 = -\frac{3^2\,13.6}{2^2}=-30.6\text{ eV}\) - For \(n=3\): \(E_3 = -\frac{3^2\,13.6}{3^2}=-13.6\text{ eV}\) - For \(n=4\): \(E_4 = -\frac{3^2\,13.6}{4^2}=-7.65\text{ eV}\) The energies of the four lowest energy states for Li+2 are: -122.4 eV, -30.6 eV, -13.6 eV, and -7.65 eV.

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