91影视

The given data is listed below as follows,

• Number of nucleons in the gold nucleus is,197
• The radius of a single nucleon is,$r=1脳{10}^{-15}\mathrm{m}$
• 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鈥檚 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:

$V=\frac{4}{3}蟺r^3$

Here, r is the radius of the nucleon.

Substitute all the values in the above expression.

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

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

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

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:

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

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

Substituting the values in the above equation.

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

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

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

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

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:

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

Here, R is the radius of the alpha particle.

Substituting the values in the above equation.

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

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

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.

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

From the coulomb鈥檚 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鈥檚 constant, and its value is$8.99脳{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2,q_1$, 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.

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

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