I saw in a textbook that gave the minimum distance for a hyperbolic trajectory in Rutherford scattering is given by

$$r_{min}= \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{2E}\left(1+\frac{1}{sin\frac{\theta}{2} }\right) $$

But I'm not sure how to derive it? Could anyone help me fill in the gaps?

My idea is to consider Energy and Momentum Conservation $$\frac{1}{2}mv_0^2 = \frac{1}{2}mv_{min}^2 + \frac{1}{4\pi\epsilon}\frac{Z_1Z_2e^2}{r_{min}} $$

$$mv_0b = mv_{min}r_{min} $$ =>$$ v_{min}=\frac{v_0b}{r_{min}}$$

which results in a quadratic equation

$$r_{min}^2-\frac{Z_1Z_2e^2}{4\pi\epsilon_0E}r_{min} - b^2 = 0 $$

Which i tried solving but couldn't get the answer. So I'm wondering if my approach is right, and if not, do let me know how do I do it!

  • $\begingroup$ Remember - hyperbolic trajectory comets going around the sun are obeying Rutherford scattering. Thinking in those terms may help. $\endgroup$
    – Jon Custer
    Sep 21, 2015 at 18:13

1 Answer 1


You're right on track but don't have enough equations. The equations which give you the solution are:

\begin{align} k&=\frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0} \tag{1}\\ E&=\frac{1}{2}m v_{min}^2+\frac{k}{r_{min}} \tag{2}\\ \frac{1}{2}m v_{min}^2&=E \frac{b^2}{r_{min}^2} \tag{3}\\ b&=\frac{k}{2 E} \cot\left(\frac{\theta}{2}\right)\tag{4} \end{align}

Equation 1 is just shorthand. Equation 2 is your second equation, but with $\frac{1}{2} mv_0^2$ replaced with $E$. Equation 3 is your fourth equation ($v_{min}=\frac{v_0 b}{r_{min}}$), but squared, multiplied by $\frac{1}{2}m$, and with $\frac{1}{2} mv_0^2$ replaced by $E$. Equation 4, finally, is the important result of Rutherford scattering relating the impact parameter to the scattering angle.

Because I'm lazy I threw it into Mathematica, but if you make these substitutions (most importantly: Getting $v_{min}$, $v_0$, and $b$ OUT of your quadratic equation), it should be easy to do by hand.

The mathematica code

 Solve[{eE == 1/2 m vmin^2 + k/rmin, 1/2 m vmin^2 == eE b^2/rmin^2, 
   b == k/(2 eE) Cot[\[Theta]/2]}, {rmin, vmin, b}], 
 Assumptions -> {eE > 0, m > 0}]

spits out, as one solution,

$$\frac{k \left(\sqrt{\csc ^2\left(\frac{\theta }{2}\right)}+1\right)}{2 \text{eE}}$$

which is exactly what you need. ("eE" is a single variable used in the mathematica code, and is not a product. This is because mathematica hates single capital letters. It represents the energy of the system, $\frac{1}{2} m v_0^2$.)


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