91Ó°ÊÓ

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

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

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,$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2}\mathrm{or}\frac{\mathrm{Ze}}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2}$(1)

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

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

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

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

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.