Question

In Rutherford scattering we noted that angular momentum is conserved. The angular
momentum of the incident α particle relative to the target nucleus is mv0b where m is the
mass, v0 is the initial velocity of the α particle, and b is the impact parameter.
Start with L = r x p and show that angular momentum is conserved, and the magnitude is
given by mv0b along the entire path of the α particle while it is scattered by the Coulomb
force from a gold nucleus

Answer 1

Let us assume the target is at very large distance more precisely at infinity and is at rest.

Generally, the force acting between target mass and alpha particle is Columb 's force(distance dependent) that

$$\overrightarrow{F}=f(r)\overrightarrow{r}$$

where,

$$f(r)=\frac{1}{4\pi\epsilon_0}\frac{q_{\alpha}q_{m}}{r^3}$$

Now, we have to show that

$$\frac{d}{dt}\overrightarrow{L}=0$$

Thus,

$$\frac{d}{dt}\overrightarrow{L}=\frac{d}{dt}(\overrightarrow{r}\times \overrightarrow{p}) \\ \implies \frac{d}{dt}(\overrightarrow{r})\times \overrightarrow{p}+\overrightarrow{r} \times\frac{d}{dt}(\overrightarrow{p})$$

But note that

$$\frac{d}{dt}(\overrightarrow{r})\times \overrightarrow{p}=m(\dot{\overrightarrow{r}}\times \dot{\overrightarrow{r}})=0$$

Now, from Newton's second law of motion , we get

$$\frac{d}{dt}(\overrightarrow{p})=f(r)\overrightarrow{r}$$

Thus,

$$\overrightarrow{r} \times\frac{d}{dt}(\overrightarrow{p})=f(r)(\overrightarrow{r} \times \overrightarrow{r} )=0$$

On combining, the above two fact, we get,

$$\dot{\overrightarrow{L}}=0$$

Thus, momentum is conserved in entire path and it is $||\dot{\overrightarrow{L}}||=mv_0b$