Derivation of Rutherford Scattering - angular momentum

In summary, the problem at hand is to determine the angular momentum magnitude at point M as an alpha particle is being scattered. The linear momentum equation and the repulsive Coulomb force are provided as relevant equations. The attempt at a solution involves finding the velocity at point M by integrating the force acting on the particle and using this to solve for angular momentum in terms of the position vector. However, the integral for finding time in terms of position proves to be difficult and a better approach is sought. The concept of angular momentum is explained as well as the components of velocity in polar coordinates.
  • #1
alexjean
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0

Homework Statement


I am trying to show that as the alpha particle is being scattered, the angular momentum magnitude at point M is [itex]m r^2 \frac{d \phi}{dt}[/itex].

Diagram:
upload_2016-1-20_10-22-57.png

Homework Equations


Linear Momentum: ##\vec L = \vec r \times \vec p ##
Repulsive Coulomb Force: ##\vec F = \frac{k 2Ze^2}{r^2} \frac{\vec r}{r}##

The Attempt at a Solution


As the alpha particle approaches the nucleus, the repulsive force begins acting on it. Changing it's velocity:
##\vec v_M = \int_{t_{-\infty}}^{t_{M}} \frac{\vec F(\phi)}{m} t \, dt + \vec v_0 ##

If I could solve that integral, I would know the velocity for any angle ##\phi##. Allowing me to solve for angular momentum in terms of ##\vec r##. However, that integral is a mess. I would need to get time, ##t## in terms of ##\phi##. I've attempted that below, but it doesn't look promising:

##t = \frac{\Delta \vec r}{\vec v} ##
## t = \int \frac{1}{\vec v} \, d\vec r ##,
## t_M = \int_{<-\infty,0>}^{\vec r_M} \frac{1}{\vec v} \, d\vec r ##
## \frac{dt}{d\vec r} = \frac{1}{\vec v} ##

Substituting into the original integral:
## \vec v_M = \int_{<-\infty,0>}^{\vec r_M} \frac{\vec F(\phi)}{m} (\int_{<-\infty,0>}^{\vec r_M} \frac{1}{\vec v} \, d\vec r) \frac{1}{\vec v} \, d\vec r + \vec v_0 ##

Most derivations of this I've found online seem to glaze over the determination of angular momentum at point M, making me think I've missed something simple. Could anyone point me toward a better approach?
 

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  • #2
alexjean said:

Homework Statement


I am trying to show that as the alpha particle is being scattered, the angular momentum magnitude at point M is [itex]m r^2 \frac{d \phi}{dt}[/itex].

Diagram:
View attachment 94534

Homework Equations


Linear Momentum: ##\vec L = \vec r \times \vec p ##
Repulsive Coulomb Force: ##\vec F = \frac{k 2Ze^2}{r^2} \frac{\vec r}{r}##

The Attempt at a Solution


As the alpha particle approaches the nucleus, the repulsive force begins acting on it. Changing it's velocity:
##\vec v_M = \int_{t_{-\infty}}^{t_{M}} \frac{\vec F(\phi)}{m} t \, dt + \vec v_0 ##

The angular momentum is ##L=\vec L = \vec r \times \vec p ##.
Is M the point of closest approach?
The momentum ,##vec p = m \vec v. In polar system of coordinates, the velocity has a radial component and a transversal one. See http://pioneer.netserv.chula.ac.th/~anopdana/211/24rtheta.pdf The vector product with the radial component is zero. The transversal component of the velocity is r dφ/dt. It is easy to get angular momentum from here.
 

1. What is the Rutherford Scattering experiment?

The Rutherford Scattering experiment is a famous physics experiment that was conducted by Ernest Rutherford in 1911. It involved firing alpha particles at a thin gold foil and observing their scattered trajectories. This experiment is important because it helped to provide evidence for the existence of the atomic nucleus and led to the development of the nuclear model of the atom.

2. How did Rutherford derive the formula for angular momentum in this experiment?

Rutherford used classical mechanics to derive the formula for angular momentum in his scattering experiment. He considered the alpha particles to be point particles and analyzed their trajectories using the principles of conservation of energy and momentum. By equating the centripetal force to the electrostatic force, he was able to derive the angular momentum formula.

3. What is the significance of angular momentum in the Rutherford Scattering experiment?

Angular momentum is a fundamental quantity in physics that is conserved in any isolated system. In the Rutherford Scattering experiment, the conservation of angular momentum played a crucial role in understanding the trajectories of the alpha particles. The formula for angular momentum also helped to determine the size and charge of the atomic nucleus.

4. How does the angle of deflection of the alpha particles relate to their angular momentum?

The angle of deflection of the alpha particles is directly proportional to their angular momentum. This means that as the angular momentum of the alpha particles increases, the angle of deflection also increases. This relationship is described by the formula for angular momentum, which shows that angular momentum is directly proportional to the product of the mass, velocity, and distance from the center of rotation.

5. What other factors besides angular momentum can affect the scattering of alpha particles in this experiment?

Besides angular momentum, the scattering of alpha particles can also be affected by the charge and mass of the particles, as well as the strength of the electrostatic force between the alpha particles and the nucleus. The thickness and material of the foil can also have an impact on the scattering patterns. Additionally, the initial velocity and direction of the alpha particles can also influence their trajectories and deflection angles.

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