Rutherford scattering differential probability function

[Uniquebum]

Homework Statement

Spray of alpha particles hit a gold membrane whose density is $\rho$ and thickness d, with a kinetic energy of $E_k$. The spray's intensity is I. Alpha particles are measured from an area of A which is set R away from the membrane. If $\theta$ is the angle between the spray hitting the membrane and the center of the area A, what is the amount of alpha particles hitting the area A in one hour when $\theta$ = 10 degrees.

Homework Equations

The propability equation for alpha particles going into an angle of $\theta$ after hitting the membrane.

$\frac{d\sigma}{d\Omega}$ = $(\frac{z*Z*e^2}{8*\pi*\epsilon*M*v^2})^2*\frac{1}{sin(\theta/2)^4}$
Where z = alpha particle's charge, Z = gold's atomic number, M = alpha particle's mass and \epsilon = vacuum permittivity

$\rho = 19.3(g/cm^3), M_{au} = 197u, Z_{au} = 79, d = 0.1*10^{-6}m, I = 10^4(Particles/Second), A = 1.0cm^2, R = 10cm$

The Attempt at a Solution

I would assume the straight forward way of doing this would be to calculate the probability for the particles to hit the angle and then multiply that with the intensity and then multiply that with 60s*60min to get the particles per one hour.

First i changed the equation to this form. ( z = 2e and i added 2*0.5 = 1 to the denominator)
$\frac{d\sigma}{d\Omega}$ = $(\frac{2e*79*e^2}{8*\pi*\epsilon*0.5*2*M*v^2})^2 * \frac{1}{sin(\theta/2)^4}$

Thus i can just throw in the kinetic energy to the denominator and make it look like
$\frac{d\sigma}{d\Omega}$ = $(\frac{79*e^3}{8*\pi*\epsilon*E_k})^2*\frac{1}{sin(\theta/2)^4}$

And after putting the numbers in i get $\approx 5.1216 \dots * 10^{-62}$
That's most likely not correct. Anyone care to help?

 

[mark726]

Thank you for your post. I can help you with calculating the number of alpha particles hitting the area A in one hour when \theta = 10 degrees.

First, let's start with the given information. We have the density of the gold membrane (\rho), its thickness (d), the kinetic energy of the alpha particles (E_k), the intensity of the spray (I), the area A, and the distance R. We also have the equation for the probability of the particles hitting at an angle \theta, which you have correctly rewritten in a simpler form.

To calculate the number of particles hitting the area A in one hour, we need to consider the following factors:

1. The number of alpha particles that pass through the membrane: This can be calculated by multiplying the intensity (I) by the time (1 hour = 3600 seconds). This gives us the total number of particles passing through the membrane in one hour.

2. The probability of the particles hitting at an angle \theta: This is given by the equation \frac{d\sigma}{d\Omega} = (\frac{79*e^3}{8*\pi*\epsilon*E_k})^2*\frac{1}{sin(\theta/2)^4}. This gives us the likelihood of the particles hitting at an angle \theta after passing through the membrane.

3. The area of A: This is given by 1.0 cm^2.

Therefore, the number of particles hitting the area A in one hour can be calculated as:
Number of particles = Intensity * Time * Probability * Area
= I * 3600 * (\frac{d\sigma}{d\Omega}) * A
= 10^4 * 3600 * 5.1216 \dots * 10^{-62} * 1.0
= 1.843776 \dots * 10^{-51}

Note: You were correct in your calculation, but you forgot to include the time factor (3600 seconds) and the area factor (1.0 cm^2) in your final equation. I hope this helps. Let me know if you have any further questions.