Rutherford Scattering Problem

In summary, the conversation discusses a Rutherford-type scattering experiment where an Alpha particle is shot towards a platinum nucleus and rebounds back along the same path. The question posed is how close the Alpha particle comes to the nucleus. The attempt at a solution involves using the equation for energy conservation to solve for the distance between the two particles, with the constant K representing Coulomb's Law and q representing the electronic charge of the particles. It is noted that K is often confused with Planck's constant, which is denoted by h and is a much smaller value. The conversation concludes with a clarification on the use of q_1 and q_2 in the equation.
  • #1
BubbaJonze
2
0

Homework Statement



An Alpha particle of a Rutherford-type scattering experiment is shot directly toward a platinum nucleus. One of the particles was apparently aimed directly at the centre of the nucleus cause it shot 180 degrees back along the same path. How close does it come to the platinum nucleus?

Homework Equations



Kinetic energy of Alpha particle: 4.0 MeV

The Attempt at a Solution


Doubt this is right, but I figured energy conservation had something to do with it.
[tex]\sum[/tex]E[tex]_{}initial[/tex]=[tex]\sum[/tex]E[tex]_{}final[/tex]

After some re-arranging, this is what I came up with:
[tex]\frac{Kq_{1}q_{2}}{r}[/tex]=E

So I figured I was solving for r, in which case i got this formula [tex]\frac{Kq_{1}q_{2}}{E}[/tex]
K= Planck's constant
q= elementary charge
E= 4.0MeV (?)

Can someone tell me if I'm anywhere near the right track? We didn't really learn about it, but my instructor stated it was a "challenger" question.
 
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  • #2
You are on the right track.

K is not Planck's constant though. [itex]K =9X10^9[/itex].

Planck's constant, on the other hand, is usually denoted by h and is [itex]6.626X10^-34[/itex]

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
 
  • #3
G01 said:
You are on the right track.

K is not Planck's constant though. [itex]K =9X10^9[/itex].

Planck's constant, on the other hand, is usually denoted by h and is [itex]6.626X10^-34[/itex]

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
Oh, that's right. I guess I got a little confused since Coulomb's Law was in my last unit and now my constants are all messed up. Thanks.
 
  • #4
Anytime.:smile:
 

Related to Rutherford Scattering Problem

1. What is the Rutherford Scattering Problem?

The Rutherford Scattering Problem is a physics problem that was first proposed by Ernest Rutherford in 1911. It involves the scattering of alpha particles (positively charged particles) off of a thin sheet of gold foil. The problem was used to study the structure of atoms and led to the discovery of the atomic nucleus.

2. What was the significance of the Rutherford Scattering Experiment?

The Rutherford Scattering Experiment was significant because it provided evidence for the existence of the atomic nucleus. Prior to this experiment, it was believed that atoms were uniform and did not have a central, positively charged nucleus. This experiment helped to refine our understanding of atomic structure.

3. How did Rutherford explain the unexpected results of the experiment?

Rutherford explained the unexpected results of the experiment by proposing the "plum pudding" model of the atom. According to this model, the atom was thought to be a uniform, positively charged sphere with negatively charged electrons scattered throughout. However, the results of the experiment showed that most of the alpha particles passed straight through the foil, indicating that the majority of the atom was actually empty space.

4. What is the mathematical equation for Rutherford Scattering?

The mathematical equation for Rutherford Scattering is known as the Rutherford Scattering Formula. It is given by: dΩ/dΩ = (ZZe²/16πε₀E)² * (1/sin⁴(θ/2)) where dΩ/dΩ is the differential cross-section, Z is the atomic number of the scattering particle, e is the elementary charge, ε₀ is the permittivity of free space, E is the kinetic energy of the scattering particle, and θ is the scattering angle.

5. What other discoveries were made as a result of the Rutherford Scattering Experiment?

Aside from the discovery of the atomic nucleus, the Rutherford Scattering Experiment also led to the discovery of isotopes. By observing the scattering patterns of different types of atoms, scientists were able to determine that atoms of the same element could have different masses, leading to the concept of isotopes. This experiment also laid the foundation for further research and advancements in the field of nuclear physics.

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