How to Find Velocity of 10 MeV Alpha Particle for Scattering Calculation?

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Homework Help Overview

The discussion revolves around calculating the cross-section for the scattering of a 10 MeV alpha particle by a gold nucleus. Participants are exploring how to determine the initial velocity of the alpha particle, which is necessary for the calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevant equations for scattering and the relationship between energy and velocity. Questions arise regarding the interpretation of the 10 MeV energy as kinetic energy and how to apply it to find the initial velocity.

Discussion Status

Several participants have provided insights into the energy relationships involved, with some suggesting that the 10 MeV should be considered purely as kinetic energy. There is ongoing clarification about the definitions of terms and the appropriate equations to use, but no consensus has been reached on the exact method to find the initial velocity.

Contextual Notes

Participants are navigating the complexities of relativistic versus classical mechanics, with some noting the rest mass of the alpha particle and its implications for the calculations. There is also mention of arithmetic challenges faced by one participant in applying the equations.

SUDOnym
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Hello
I actually wanted to posted this in the "Homework" section but it is currently working for me..

The problem is:

Calculate the cross-section for the scattering of a 10 MeV alpha particle by a gold nucleus _{79}^{197}Au through an angle greater than (a) 10 degrees (b) 20 degrees c 30 degrees.

My answer:

I know the relevant equation is:

\sigma=\pi(\frac{Zze^{2}}{4\pi\epsilon_{0}mv_{0}^{2}}\cot\frac{\theta}{2})^{2}

so its really just a plug and chug kind of problem... only thing I can't figure out is how to get the value for v_0... i know that is hidden somewhere in the fact that it is a 10 Mev particle and I assume E=mc^2 plays a roll too... but trying the following does not provide me with a useful value of v_o:

E=mc^{2}+\frac{1}{2}mv_{0}^{2}

neither does:

E=\frac{1}{2}mv_{0}^{2}

so my question is, how do I find the value of v_0 given that I know its an alpha particlee (so I know its mass) and also that I know its 10Mev particle?

Many Thanks!
 
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What is "v_0" ?
 
v_0 is the initial velocity that the alpha particle has before any interaction with the gold nucleus..
 
The correct relativistic relationship between energy and momentum is
E^2 = p^2 c^2 + m^2 c^4

Here m is the invariant mass (in the rest frame) and p =mv.
From this you can find p and then v.
 
Thanks! to clarify though, the energy E, that I want to use will be the 10MeV...?
 
I think the 10 MeV should be just the kinetic energy. The rest energy of a proton is about 900 MeV so the 10 MeV cannot include the rest mass of the alpha particle.
The kinetic energy is
E^2 - m^2 c^4 = p^2 c^2

Actually at this energy the classic approximation should work quite well as KE<<E.
 
thanks, nasu. E=1/2mv^2 was my initial try but I kept making arithmetic mistakes which is why I came on here... took me at least two hours to get the correct answer for the cross-section!
 
SUDOnym said:
it is a 10 Mev particle and I assume E=mc^2 plays a roll too...
Nope! Rest mass of \alpha-particle is 3.7GeV, so if you are speaking about 10MeV \alpha, it is its pure kinetic energy, and you have pretty non-relativistic case.
 

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