Question

relation between distance of closest approach and impact parameter

Answer 1

r_0 = \frac{k}{2E} \left( 1 + \sqrt{1 + \left(\frac{2Eb}{k}\right)^2} \right) \text{ or } b^2 = r_0^2 - \frac{k}{E}r_0 \text{ where } k = \frac{Z_1 Z_2 e^2}{4\pi\epsilon_0} \text{ and } E = \frac{1}{2}mv^2

Answer 2

In Rutherford scattering, the trajectory of an alpha particle (charge $Z_1 e$) approaching a nucleus (charge $Z_2 e$) is a hyperbola due to the repulsive Coulomb force. The initial kinetic energy of the alpha particle is $E = \frac{1}{2}mv^2$, where $m$ is its mass and $v$ is its initial speed far from the nucleus. The interaction potential is $V(r) = \frac{Z_1 Z_2 e^2}{4\pi\epsilon_0 r} = \frac{k}{r}$, where $k = \frac{Z_1 Z_2 e^2}{4\pi\epsilon_0}$.

We can relate the distance of closest approach ($r_0$) and the impact parameter ($b$) using the conservation of energy and angular momentum.

1. Conservation of Energy:

The total energy of the alpha particle is conserved. Far from the nucleus ($r \to \infty$), the potential energy is zero, and the total energy is just the initial kinetic energy $E = \frac{1}{2}mv^2$.

At the point of closest approach ($r = r_0$), the velocity ($v_0$) is purely tangential (perpendicular to the position vector $\vec{r}_0$), and the total energy is the sum of kinetic and potential energy:

$E = \frac{1}{2}mv_0^2 + \frac{k}{r_0}$

2. Conservation of Angular Momentum:

The angular momentum of the alpha particle about the nucleus is conserved. Far from the nucleus, the initial angular momentum is $L = mvb$, where $b$ is the impact parameter.

At the point of closest approach ($r = r_0$), the velocity $v_0$ is perpendicular to the position vector $r_0$, so the angular momentum is $L = mv_0 r_0$.

Equating the initial and final angular momentum:

$mvb = mv_0 r_0 \implies v_0 = \frac{vb}{r_0}$

3. Substitute $v_0$ into the energy equation:

$E = \frac{1}{2}m\left(\frac{vb}{r_0}\right)^2 + \frac{k}{r_0}$

$E = \frac{1}{2}m\frac{v^2b^2}{r_0^2} + \frac{k}{r_0}$

Since $E = \frac{1}{2}mv^2$, we can write $\frac{1}{2}mv^2 b^2 = Eb^2$.

$E = \frac{Eb^2}{r_0^2} + \frac{k}{r_0}$

Multiply the equation by $r_0^2$:

$Er_0^2 = Eb^2 + kr_0$

Rearrange into a quadratic equation in $r_0$:

$Er_0^2 - kr_0 - Eb^2 = 0$

4. Solve the quadratic equation for $r_0$:

Using the quadratic formula $r_0 = \frac{-(-k) \pm \sqrt{(-k)^2 - 4(E)(-Eb^2)}}{2E}$:

$r_0 = \frac{k \pm \sqrt{k^2 + 4E^2b^2}}{2E}$

Since $r_0$ must be a positive distance, we take the positive root:

$r_0 = \frac{k + \sqrt{k^2 + 4E^2b^2}}{2E}$

$r_0 = \frac{k}{2E} + \frac{\sqrt{k^2 + (2Eb)^2}}{2E}$

This equation gives the distance of closest approach $r_0$ as a function of the impact parameter $b$ and the initial kinetic energy $E$.

We can also rearrange this equation to express $b$ in terms of $r_0$:

$2Er_0 = k + \sqrt{k^2 + 4E^2b^2}$

$2Er_0 - k = \sqrt{k^2 + 4E^2b^2}$

Squaring both sides:

$(2Er_0 - k)^2 = k^2 + 4E^2b^2$

$4E^2r_0^2 - 4Er_0k + k^2 = k^2 + 4E^2b^2$

$4E^2r_0^2 - 4Er_0k = 4E^2b^2$

Divide by $4E^2$:

$r_0^2 - \frac{k}{E}r_0 = b^2$

$b = \sqrt{r_0^2 - \frac{k}{E}r_0}$

Let $r_{0,min} = \frac{k}{E} = \frac{Z_1 Z_2 e^2}{4\pi\epsilon_0 E}$. This is the distance of closest approach for a head-on collision ($b=0$). The relation can also be written as:

$b^2 = r_0^2 - r_{0,min} r_0$

The question asks for the relation between the distance of closest approach and the impact parameter. The derived equations provide this relationship.