Question

(a) To determine the repulsive energy between the two electrons in helium.

(b) To determine the distance of electrons that would have to be separated.

(c) To compare distance with approximate orbit radius in Z=2hydrogen like atom.

Short answer

(a)${\mathrm{E}}_{\mathrm{R}}=29.8\mathrm{eV}$.

(b)To produce the required repulsive energy, tile electrons would have to be separated by a distance of $\mathrm{r}=4.8\times {10}^{-11}\mathrm{m}$.

(c) The ratio of the distances is nearly 2.

Step-by-step solution

Use Formula for energy of an electron.

Formula

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\left(-13.6\mathrm{eV}\right){\left(\frac{Z}{n}\right)}^{\mathbf{2}}$ …(1)

The energy E_n of an electron in a hydrogen-like atom with Z number of protons is:

Here n is the principal quantum number of the electron.

Calculate how much energy to use to remove the second election

The energy needed to remove the first electron from helium is 24.6eV.

Equation (1) is used to find how much energy to use to remove the second election, with n being 1 and Z being 2 .

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\left(-13.6\mathrm{eV}\right){\left(\frac{\mathrm{Z}}{\mathrm{n}}\right)}^2 \\ {\mathrm{E}}_1=\left(-13.6\mathrm{eV}\right){\left(\frac{2}{1}\right)}^2 \\ {\mathrm{E}}_1=-54.4\mathrm{eV}. \end{gathered}$$

So since the second electron is bound with energy of 54.4eV that's how much energy needs to be added to remove that electron. If there was no repulsive energy between the two electrons, both of them would need 54.4eV in order to be removed from the atom. So that can be expressed with:

54.4eV+54.4eV -E_R=24.6 eV + 54.4eV

Here E_R is the repulsive energy between the electrons, which lessens the amount of energy that needs to be added to completely ionize the atom. That is then simplified and solved for E_R :

$\begin{array}{rcl}54.4\mathrm{eV}+54.4\mathrm{eV}-{\mathrm{E}}_{\mathrm{R}}& =& 24.6\mathrm{eV}+54.4\mathrm{eV}\\ 108.8\mathrm{eV}-{\mathrm{E}}_{\mathrm{R}}& =& 79\mathrm{eV}\\ {\mathrm{E}}_{\mathrm{R}}& =& 29.8\mathrm{eV}\end{array}$

The repulsive energy between the two electrons in helium is${\mathrm{E}}_{\mathrm{R}}=29.8\mathrm{eV}$

Use formula of required repulsive energy

Formula

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{r}}$

Calculate how far apart the electrons would be in order to have that much repulsive energy.

To find how far apart the electrons would be in order to have that much repulsive energy, its first converted to Joules.

The repulsive energy between the two electrons in helium is E_R=29.8eV

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{n}}& =& \left(29.8\mathrm{eV}\right)\left(\frac{1.6\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}\right)\\ {\mathrm{E}}_{\mathrm{R}}& =& 4.768\times {10}^{-18}\mathrm{J}\end{array}$

That is then used as the E in equation (3).

$\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{\mathrm{r}}$

Rearrange above equation for r

$r=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{E}$

Substitute$1.6\times {10}^{-19}\mathrm{C}$ for Q₁,Q₂,$8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2$ fordata-custom-editor="chemistry" ${\mathrm{\epsilon }}_0$ and$4.768\times {10}^{-18}\mathrm{J}$ for E.

$$\begin{gathered} \mathrm{r}=\frac{1}{4\mathrm{\pi }\left(8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\right)}\frac{{\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{\left(4.768\times {10}^{-18}\mathrm{J}\right)} \\ \mathrm{r}=4.83\times {10}^{-11}\mathrm{m} \end{gathered}$$

To produce the required repulsive energy, tile electrons would have to be separated by a distance of$\mathrm{r}=4.83\times {10}^{-11}\mathrm{m}$.

Use Formula for radius of electron.

Formula:

${\mathbf{r}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{a}}_{\mathbf{0}}}{\mathbf{Z}}$

Calculate distance between electrons compares with the approximate radius of the atom.

From equation (2) the approximate radius r_n of a hydrogen-like atom with Z number of protons is.

${\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{a}}_0}{\mathrm{Z}}$

Substitute 1 for n, 2 for Z and$5.29\times {10}^{-11}\mathrm{m}$for a₀

$$\begin{gathered} {\mathrm{r}}_1=\frac{{\left(1\right)}^2\left(5.29\times {10}^{-11}\mathrm{m}\right)}{2} \\ {\mathrm{r}}_1=2.645\times {10}^{-11}\mathrm{m} \end{gathered}$$

The ratio of the two distances is then

$\begin{array}{rcl}\frac{4.83\times {10}^{-11}m}{2.645\times {10}^{-11}m}& =& 1.825\\ & \approx & 2\end{array}$

Hence the ratio of the two distances is, 2.