Question

The radius of a gold nucleus (Z = 79) is about 7·0×10-10 m. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?

Answer 1

Answer:

Explanation:

Given:

Radius of gold nucleus=7.0× 10-15m=r

Charge on the nucleus=79× 1.6× 10-19C=q

(a)

Consider a gaussian spherical surface of radius of the nucleus=r

By symmetry all points on this surface are equivalent and electric field at all points have same magnitude and is in radial direction

Therefore flux passing through this surface is given by

…(i)

We know that,

By Gauss’s law, flux of net electric field (E⃗ ) through a closed surface S equals the net charge enclosed (qin) by the surface divided by ϵ0

Charge enclosed by this sphere =total charge on the nucleus=q

Using gauss’s law and eqn.(i)

Putting the values of r,q and ϵ0

N/C

Therefore the strength of electric field at the surface of nucleus is 2.31× 1021N/C

(b)for calculating electric field at middle point of radius consider similar gaussian surface with radius half of that of nucleus=r/2

Charge enclosed by volume =Q

Charge enclosed by volume Q’

We get

Using gauss law and eqn.(i)

N/C

N/C

Therefore electric field at middle point of radius of nucleus is given by 1.16× 1021N/C

We know that when electric charge is given to a conductor it comes on its surface

But nucleons are bound by strong nuclear force inside nucleus which holds them and prevents them from coming out of conductor.

Therefore it is justified to assume that electric charge is uniformly distributed in its entire volume.

Answer 2

Gold’s atomic number is Z = 79. This means that inside the nucleus, there are 79 protons. So, the total charge is $Q=79 e=1.264 \times 10-17 C$ The nucleus’ radius is $r=7 \times 10-15 \mathrm{m}$.

Explanation:

(a) Assume a spherical surface of the nucleus with the radius, r. All the point by symmetry on the surface are equivalent. The electric fields have same magnitude. Hence, through this surface, the passage of flux is shown by

$\phi=\oint \vec{E} \cdot d \vec{S}=E \oint d \vec{S}=E .4 \pi r^{2}$

We know that,

$\oint \vec{E} \cdot d \vec{S}=\frac{q_{i n}}{\epsilon_{0}}$

Applying gauss’s law in eqn.(i),

$\begin{aligned}&E .4 \pi r^{2}=\frac{q}{\epsilon_{0}}\\&E=\frac{q}{4 \pi r^{2} \epsilon_{0}}\\&E=79 \times 1.6 \times \frac{10^{-19}}{4 \times 3.14 \times\left(7 \times 10^{-10}\right)^{2} \times 8.85 \times 10^{-12}}\\&\text { so, } E=2.31 \times 10^{21}\end{aligned}$

So, at the nucleus surface, the electric field strength is 2.31× 1021N/C.

(b)

Charge = $\frac{4}{3} \pi r^{3}=\mathrm{Q}$

We get

$Q^{\prime}=\frac{Q}{8}$

Using gauss law and eqn.(i)

$\begin{aligned}&E \cdot 4 \pi r^{2}=\frac{Q}{8 \epsilon_{0}}\\&E=\frac{Q}{32 \pi r^{2} \epsilon_{0}}\\&E=2.31 \times \frac{10^{21}}{8} \mathrm{N} / \mathrm{C}\\&E=1.16 \times 10^{21} \mathrm{N} / \mathrm{c}\end{aligned}$

So, the electric field in the nucleus middle point is given by $1.16 \times 10^{21} \mathrm{N} / \mathrm{C}$

Hence, the assumption can be justified that there is an uniform distribution of electric charge in its volume.