Question

A doubly ionized lithium atom(Li⁺⁺) is one that has hadtwo of its three electrons removed. The energy levels of the remaining single-electron ion are closely related to those of the hydrogen atom. The nuclear charge for lithium is+3e instead of just+e. How are the energy levels related to those of hydrogen? How is the radius of the ion in the ground level related to that of the hydrogen atom ? Explain.

Short answer

The energy of ionised atom is 9 times the energy of the neutral hydrogen atom and the radius of the ionised atom is one-third of the neutral hydrogen atom.

Step-by-step solution

Important Concepts

We have the following relations from Bohr’s atomic model

Energy in n^thorbit

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\frac{\mathbf{h}\mathbf{c}\mathbf{R}}{{\mathbf{n}}^{\mathbf{2}}}$

Where h is the plank’s constant, c is the speed of light and R is the Rydberg Constant.

Radius of the n^thorbit

${\mathit{r}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{n}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{{\mathbf{\pi me}}^{\mathbf{2}}}$

And R is given by

$\mathit{R}\mathbf{=}\frac{\mathbf{m}{\mathbf{e}}^{\mathbf{4}}}{\mathbf{8}{\mathbf{\epsilon }}_{\mathbf{0}}^{\mathbf{4}}{\mathbf{h}}^{\mathbf{3}}\mathbf{c}}$

Application

The doubly ionised lithium ion$L{i}^{++}$is considered as a hydrogen-like atom with Z=3

We replace$e^2$by$3e^2$everywhere.

For Radius,

The radius of the ion in the ground state is given by

$$\begin{gathered} r_1\left(L{i}^{++}\right)=\frac{{\epsilon }_0{\left(1\right)}^2{h}^2}{\mathrm{\pi m}3{\mathrm{e}}^2}=\frac{1}{3}\frac{{\epsilon }_0{h}^2}{{\mathrm{\pi me}}^2} \\ {r}_1\left(L{i}^{++}\right)=\frac{1}{3}{r}_1\left(H\right) \end{gathered}$$

For energy input the expression of the constant R into the energy expression,

We get

$E_n=-\frac{m{Z}^2{e}^2}{8{\epsilon }_0^2{h}^2{n}^2}$

For doubly ionised lithium

$$\begin{gathered} E_n\left(L{i}^{++}\right)=\frac{m{\left(3\right)}^2{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}=9-\left(\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}\right) \\ {E}_n\left(L{i}^{++}\right)=9E_n\left(H\right) \end{gathered}$$

Hence,The energy of ionised atom is 9 times the energy of the neutral hydrogen atom and the radius of the ionised atom is one-third of the neutral hydrogen atom.