Question

Buckminsterfullerine is a molecule of 60 carbon atoms arranged

like the stitching on a soccer-ball. It may be approximated as a conducting spherical shell of radius $\mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}\mathbf{°}$. A nearby electron would be attracted, according to Prob. 3.9, so it is not surprising that the ion ${\mathit{C}}_{\mathbf{60}}^{\mathbf{-}}$exists. (Imagine that the electron on average-smears itself out uniformly over the surface.) But how about a second electron? At large distances it would be repelled by the ion, obviously, but at a certain distance r (from the center), the net force is zero, and closer than this it would be attracted. So an electron with enough energy to get in that close should bind.

(a) Find r, in $\mathit{A}\mathbf{°}$. [You'll have to do it numerically.]

(b) How much energy (in electron volts) would it take to push an electron in (from

infinity) to the point r? [Incidentally, the ${\mathit{C}}_{\mathbf{60}}^{\mathbf{-}}$ion has been observed.]

Short answer

Answer:

(a) The distance at which the force on an electron from a ${\mathit{C}}_{\mathbf{60}}^{\mathbf{-}}$ion is zero is $\mathbf{5}\mathbf{.}\mathbf{663}\mathbf{}\mathit{A}\mathbf{°}$.

(b) The work done to bring an electron from infinity to the distance where the force on it is zero from a ${\mathit{C}}_{\mathbf{60}}^{\mathbf{-}}$ion is $\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{}\mathit{e}\mathit{V}$.

Step-by-step solution

Given data

There is a molecule of 60 carbon atoms approximated as a conducting spherical

shell of radius $\mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}\mathbf{°}$. A nearby electron on average-smears itself out uniformly over the surface.

Forces on a charge

The force of attraction on a point charge q at a distance afrom a neutral conducting sphere of radius R is

$\mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\left(r^2-R^2\right)}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free space.

The force of repulsion between two similar point charges q at a distance a from each other is

$\mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Distance of zero force on a charge from a negative ion

From equations (1) and (2), the net force on the incoming electron is

$$\begin{gathered} \mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\left(r^2-R^2\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right) \end{gathered}$$

For the net force to be zero,

$$\begin{gathered} {\mathit{r}}_{\mathbf{0}}\mathbf{-}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{r}}_{\mathbf{0}}{\mathbf{(}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathit{R}}^{\mathbf{3}}\mathbf{(}\mathbf{2}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)} \end{gathered}$$

Solving this in Mathematica,

${\mathit{r}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{663}\mathbf{}\mathit{A}\mathbf{°}$

Thus, the distance at which force is zero is $\mathbf{5}\mathbf{.}\mathbf{663}\mathbf{}\mathit{A}\mathbf{°}$.

Work done to bring a charge from infinity to zero force distance from a negative ion

The work done to bring the electron from infinity to ${\mathit{r}}_{\mathbf{0}}$is

$$\begin{gathered} \mathit{W}\mathbf{=}\mathbf{-}\underset{\mathbf{\infty }}{\overset{{\mathbf{r}}_{\mathbf{0}}}{\mathbf{\int }}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right)\mathit{d}\mathit{r} \\ \mathbf{=}\mathbf{-}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\underset{\mathbf{\infty }}{\overset{{\mathbf{r}}_{\mathbf{0}}}{\mathbf{\int }}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right)\mathit{d}\mathit{r} \end{gathered}$$

Substitute $\mathit{x}\mathbf{=}\frac{\mathbf{r}}{\mathbf{R}}$

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\underset{{\mathbf{x}}_{\mathbf{0}}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{)}\mathit{d}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\left[}\frac{\mathbf{1}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{0}}\mathbf{-}\mathbf{2}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{3}}}{\mathbf{2}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{)}}\mathbf{\right]} \end{gathered}$$

Substitute value of ${\mathit{r}}_{\mathbf{0}}$to get

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{R}} \end{gathered}$$

Substitute

$$\begin{gathered} \mathit{q}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathit{C} \\ \frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\frac{\mathbf{N}\mathbf{·}{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}} \\ \mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}\mathbf{°} \end{gathered}$$

and get,

$\mathit{W}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{}\mathit{e}\mathit{V}$

Thus, the work done is $\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{}\mathit{e}\mathit{V}$.