I did an experiment on Rutherford scattering , finding the number of counts at an angle $\theta$. The problem is that I had a single sensor i.e. I measured the counts at a single position , and the scattering occurs over a cone. I was thinking of integrating my obtained function, $f(\theta)$ over the area of a hoop and so obtaining :

$N(\theta) = \int_{0}^{2\pi}f(\theta)d\phi$

Where $d\phi$ is the angle subtended by part of the hoop. Therefore I get just

$N(\theta) = 2\pi f(\theta)$

Because surely

$f(\theta) = \frac{N(\theta)}{d\phi}$


1 Answer 1


The angle $d\phi$ that your detector subtends (out of a possible $2\pi$) changes with $r$; if the circumference of the "hoop" is $2\pi r \sin\theta$, and the dimension of the detector perpendicular to the $\theta$ direction is $d$, then

$$d\phi = \frac{d}{2\pi R \sin\theta}$$

where $R$ is the distance to the detector.

After that, your argument works - so the fraction $F$ of counts you observe is

$$F = \frac{d}{2\pi R\sin\theta}$$

and the conversion from $f(\theta)$ to $N(\theta)$ is

$$N(\theta) = \frac{f(\theta)}{F}=\frac{2\pi R\sin\theta}{d}f(\theta)$$

The term $\frac{2\pi R}{d}$ is something that will be constant for the apparatus; you are left with a $\sin\theta$ scaling term. The approximation will obviously break down when the detector dimension gets large compared to the size of the cone... i.e. for very small values of $\theta$ you have to be careful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.