Question

The nucleus of a plutonium$\mathbf{-}\mathbf{239}$atom containsrole="math" localid="1657279163265" $\mathbf{94}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{t}\mathit{o}\mathit{n}\mathit{s}\mathbf{.}$. Assume that the nucleus is a sphere with radius $\mathbf{6}\mathbf{.}\mathbf{64}\mathbf{}\mathbf{fm}$ and with the charge of the protons uniformly spread through the sphere. At the surface of the nucleus, what are the (a) magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons?

Short answer

a) The magnitude of the electric field produced by the protons at the surface of the nucleus is$3.07\times {10}^{2}1\hspace{0.33em}\mathrm{N}/\mathrm{C}$

b) The direction of the electric field produced by the protons at the surface of the nucleus is radially outward.

Step-by-step solution

The given data

a) The nucleus has a radius, $\mathrm{R}=6.64\mathrm{fm}\mathrm{or}6.64\times {10}^{-15}\mathrm{m}$

b) The number of protons in the nucleus of the plutonium, $Z=94$

Understanding the concept of electric field 

Since the nucleus is treated as a sphere with uniform surface charge distribution, the electric field at the surface is the same as it would be if the charge were all at the center. Thus, using the concept of the electric field, we can get the required value of the electric field and its direction.

Formula:

The magnitude of the electric field at the nucleus surface,$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\mathrm{or}\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}$(1)

where R is the nuclear radius role="math" localid="1657279658900" $\mathrm{q}(\mathrm{or}\mathrm{Ze})$,is the total charge of the nucleus.

a) Calculation of the magnitude of the electric field

Using the given data in equation (i), we can get the magnitude of the electric field as given:

$$\begin{gathered} \mathrm{E}=\frac{8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times 94\times 1.6\times {10}^{-19}\mathrm{C}}{{\left(6.64\times {10}^{-15}\mathrm{m}\right)}^2} \\ =3.07\times {10}^{21}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $3.07\times {10}^{21}\hspace{0.33em}\mathrm{N}/\mathrm{C}$

b) Calculation of the direction of the electric field

The direction of electric field lines is radially outward for a positive charge, and radially inward for a negative charge. Since the charge of the protons is normal to the surface, the field is also normal to the given surface.

Hence, the electric field is radially outward in direction. The field lines of our nucleus are shown on the right.