91影视

An alpha particle projected to the aluminum nucleus stops momentarily at a distance.

Rutherford scattering experiment shows a beam of high-energy streams of伪-particles from a radioactive source directed at a thin sheet (100 nm thickness) of gold. To study the deflection caused to the伪-particles, he placed a fluorescent zinc sulfide screen around the thin gold foil.

The distance of the closest approach is defined as the minimum distance of the charged particle from the nucleus at which initial kinetic energy is the same as the potential energy of the nucleus.

Formulae:

The potential energy of the system of charged particles,$P.E=\frac{k{q}_1{q}_2}{r}鈥�....鈥�\left(1\right)$

The radius of an atom of mass number, A in a nucleus, $r=r_0{A}^{1/3}路路路路路路路\left(2\right)where{r}_0=1.2fm$

Using equation (2), we estimate the nuclear radii of the alpha particle and aluminum as follows:

$$\begin{gathered} r_{伪}=\left(1.2脳{10}^{-15}m\right){\left(4\right)}^{1/3} \\ =1.90脳{10}^{-15}m \end{gathered}$$

$$\begin{gathered} r_{Al}=\left(1.2脳{10}^{-15}m\right){\left(27\right)}^{1/3} \\ =3.69脳{10}^{-15}m \end{gathered}$$

Thus, the distancebetween the centers of the nuclei when their surfaces touch is given by:

$$\begin{gathered} \mathrm{r}={\mathrm{r}}_{\mathrm{伪}}+{\mathrm{r}}_{\mathrm{Al}} \\ =1.90脳{10}^{-15}\mathrm{m}+3.60脳{10}^{-15}\mathrm{m} \\ =5.50脳{10}^{-15}\mathrm{m} \end{gathered}$$

Now, from the condition of closest distance of approach, the kinetic energy is equal to potential energy of the system and thus is given using equation (i) as follows;

$$\begin{gathered} K.E=\frac{\left(9脳{10}^{9}N.m^2/C^2\right)\left(2脳1.6脳{10}^{-19}C\right)\left(13脳1.6脳{10}^{-19}C\right)}{5.50脳{10}^{-15}m} \\ =1.09脳{10}^{-12}J \\ =6.79脳{10}^{6}eV \\ =6.79MeV \end{gathered}$$

Hence, the kinetic energy value is 6.79 MeV.