A gold nucleus contains 197nucleons (79protons and 118neutrons) packed tightly against each other. A single nucleon (proton or neutron) has a radius of about$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathbf{m}$. Remember that the volume of a sphere is$\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi r}}^{\mathbf{3}}$. (a) Calculate the approximate radius of the gold nucleus. (b) Calculate the approximate radius of the alpha particle, which consists of 4nucleons, 2protons and 2neutrons. (c) What kinetic energy must alpha particles have in order to make contact with a gold nucleus?

Rutherford correctly predicted the angular distribution for 10 MeV(kinetic energy) alpha particles colliding with gold nuclei. He was lucky: if the alpha particle had been able to touch the gold nucleus, the strong interaction would have been involved and the angular distribution would have deviated from that predicted by Rutherford, which was based solely on electric interactions.

The given data is listed below as follows,

• Number of nucleons in the gold nucleus is,197
• The radius of a single nucleon is,$r=1\times {10}^{-15}\mathrm{m}$
• The 197 nucleons consist of 118 neutrons and 79 protons.

The alpha particle consists of 4 nucleons and 2 protons, and two neutrons

The volume is described as a three-dimensional quality used to measure a solid object’s capacity. It is described as four third of the product of the cube of the radius for a sphere and the pi.

The kinetic energy is described as half of the product of the mass and the square of the velocity of an object.

The equation of the volume of the nucleon is expressed as:

$V=\frac{4}{3}\pi r^3$

Here, r is the radius of the nucleon.

Substitute all the values in the above expression.

$$\begin{gathered} V=\frac{4}{3}\times 3.14\times {\left({10}^{-15}\mathrm{m}\right)}^3 \\ =4.19\times {10}^{-45}{\mathrm{m}}^3 \end{gathered}$$

As there are 197 nucleons, so the total volume of the nucleon is as follows,

$$\begin{gathered} 197\times V=197\times 4.19\times {10}^{-45}{\mathrm{m}}^3 \\ =8.25\times {10}^{-43}{\mathrm{m}}^3 \end{gathered}$$

The total volume of the nucleon is the total volume of the gold nucleus. So, the equation of the volume of the gold nucleus is expressed as:

$$\begin{gathered} V=\frac{4}{3}\pi R_1^3 \\ {R}_1=\sqrt[3]{\frac{3V}{4\pi }} \end{gathered}$$

Here,$R_1$is the radius of the gold nucleus.

Substituting the values in the above equation.

$$\begin{gathered} R=\sqrt[3]{\frac{3\times 8.25\times {10}^{-43}{\mathrm{m}}^3}{4\pi }} \\ =5.81\times {10}^{-15}\mathrm{m} \end{gathered}$$

Thus, the approximate radius of the gold nucleus is$5.81\times {10}^{-15}\mathrm{m}$.

As there are four nucleons, so the total volume of the nucleon is as follows,

$$\begin{gathered} 4\times V=4\times 4.19\times {10}^{-45}{\mathrm{m}}^3 \\ =1.676\times {10}^{-44}{\mathrm{m}}^3 \end{gathered}$$

The total volume of the nucleon is the total volume of the alpha particle. So, the equation of the volume of the alpha particle is expressed as:

$$\begin{gathered} V=\frac{4}{3}\pi R^3 \\ R=\sqrt[3]{\frac{3V}{4\pi }} \end{gathered}$$

Here, R is the radius of the alpha particle.

Substituting the values in the above equation.

$$\begin{gathered} R=\sqrt[3]{\frac{3\times 1.676\times {10}^{-44}{\mathrm{m}}^3}{4\pi }} \\ =1.59\times {10}^{-15}\mathrm{m} \end{gathered}$$

Thus, the approximate radius of the alpha particle is$1.59\times {10}^{-15}\mathrm{m}$.

The equation of the distance between the alpha particle and the gold nucleus is expressed as:

$r=r_a+r_g$

Here, ris the distance between the alpha particle and the gold nucleus,$r_a$is the radius of the alpha particle and$r_g$is the radius of the gold nucleus.

Substitute the values in the above equation.

$$\begin{gathered} r=1.59\times {10}^{-15}\mathrm{m}+5.81\times {10}^{-15}\mathrm{m} \\ =7.4\times {10}^{-15}\mathrm{m} \end{gathered}$$

From the coulomb’s law, the equation of the kinetic energy of the alpha particles is expressed as:

$K.E.=k\frac{q_1{q}_2}{r}$

Here, ris Coulomb’s constant, and its value is$8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2,q_1$, is a charge of the gold nucleus,$q_2$is a charge of the alpha particle and r is the distance between the alpha particle and the gold nucleus.

Substitute all the values in the above expression.

$$\begin{gathered} K.E.=8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{79\times 1.602\times {10}^{-19}\mathrm{C}\times 2\times 1.602\times {10}^{-19}\mathrm{C}}{\left(7.4\times {10}^{-15}\mathrm{m}\right)} \\ =8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{4.054\times {10}^{-36}{\mathrm{C}}^2}{\left(7.4\times {10}^{-15}\mathrm{m}\right)} \\ =8.99\times {10}^9\times 5.47\times {10}^{-22}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times {\mathrm{C}}^2/\mathrm{m} \\ =4.92\times {10}^{-12}\mathrm{N}.\mathrm{m} \\ =4.92\times {10}^{-12}\mathrm{N}.\mathrm{m} \\ =4.92\times {10}^{-12}\mathrm{N}.\mathrm{m}\times \left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =4.92\times {10}^{-12}\mathrm{J} \end{gathered}$$

Thus, the kinetic energy the alpha particles must have to contact a gold nucleus is$4.92\times {10}^{-12}\mathrm{J}$