91Ó°ÊÓ

Consider the following graph for Rutherford experiment, we need to find the scattering angle $\mathrm{θ}$. Assume that the impulse is a small fraction of the particle's incoming horizontal momentum, so the scattering angle should be small, and given approximately by:

$\mathrm{θ}={\tan }^{-1}\left(\frac{l}{p}\right)$…â¶Ä¦..(1)

In Rutherford experiment, the impulse is the change in momentum that a force produces over a given time, that is:

$l=∫F_{âŠ\yen }dt$…â¶Ä¦(2)

Where, $F_{âŠ\yen }$is the perpendicular component of the force.

The force is given by:

$F=\frac{1}{4\mathrm{π}{∈}_0}\frac{q_1{q}_2}{r^2}\hat{r}$

So, the perpendicular component is given by:

$F_{âŠ\yen }=\frac{1}{4\mathrm{Ï€}{∈}_0}\frac{q_1{q}_2}{r^2}\cos \left(\mathrm{Ï•}\right)$

Where,

$\cos \left(\mathrm{Ï•}\right)=\frac{b}{r}$

Thus,

$F_{âŠ\yen }=\frac{1}{4\mathrm{Ï€}{∈}_0}\frac{q_1{q}_{2}b}{r^3}$

Substitute into (2) to get:

$l=\frac{q_1{q}_{2}b}{4\mathrm{Ï€}{∈}_0}{∫}_{-∞}^{∞}\frac{dt}{r^3}$…â¶Ä¦(3)

We need to convert this integral to be over $r$, using:

$v=\frac{dx}{dt}=\sqrt{\frac{2E}{m}}$

From the graph we have:

$x=\sqrt{r^2-b^2}$

So,

$$\begin{gathered} dx=\frac{rdr}{\sqrt{r^2-b^2}} \\ dx=\sqrt{\frac{2E}{m}}dt \end{gathered}$$

Substitute into (3) we get:

$l=\frac{2q_1{q}_{2}b}{4\mathrm{π}{∈}_0\sqrt{2Elm}}{∫}_b^{∞}\frac{dr}{r^2\sqrt{r^2-b^2}}$

Where $r$the integral is over the range of $r$from its closest approach$b$,. To do theintegral I used the integral calculator, so we get:

$$\begin{gathered} {∫}_b^{∞}\frac{dr}{\sqrt{r^2-b^2}}={\overline{)\frac{\sqrt{r^2-b^2}}{b^{2}r}}}_b^{∞} \\ {∫}_b^{∞}\frac{dr}{\sqrt{r^2-b^2}}=\frac{1}{b^2} \end{gathered}$$

Substitute into (4), so we get:

$l=\frac{q_1{q}_2}{4\mathrm{π}{∈}_{0}b\sqrt{2Elm}}$

Now substitute into equation (1), note that the momentum in terms of the energy is $p=\sqrt{2mE}$, so we get:

localid="1655982043823" $\mathrm{θ}={\tan }^{-1}\left(\frac{q_1{q}_2}{4\mathrm{π}{∈}_{0}bE}\right)$

The impact parameter in problem 11.1, is:

localid="1655982050659" $b=\frac{q_1{q}_2}{8\mathrm{π}{∈}_{0}E}cot\left(\frac{\mathrm{θ}}{2}\right)$

(Or)

localid="1655982055887" $\tan \left(\frac{\mathrm{θ}}{2}\right)=\frac{q_1{q}_2}{8\mathrm{π}{∈}_{0}bE}$

For small angle we have localid="1655982062799" $\frac{\mathrm{θ}}{2}≈\frac{\mathrm{θ}}{2}$so we get:

localid="1655982068822" $$\begin{gathered} \frac{\mathrm{θ}}{2}≈\frac{q_1{q}_2}{8\mathrm{π}{∈}_{0}bE} \\ \frac{\mathrm{θ}}{2}≈\frac{q_1{q}_2}{4\mathrm{π}{∈}_{0}bE} \end{gathered}$$

Which is the same as in equation (5) for small angle localid="1655982077648" $\mathrm{θ}$in (5).