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Homework Help: Very basic (?) particle/nuclear physics

  1. Jul 8, 2010 #1
    Hi! I am taking a course called Modern Physics and I have this small issue when it comes to particle physics.

    1. The problem statement, all variables and given/known data

    "An [tex]\alpha[/tex]-particle collides with a silicon nucleus, 29Si, and a nuclear reaction takes place. One proton is emitted. What is the reaction?
    Hint: 12Mg,13Al, 14Si,15P, 16S are "available" nuclei."

    2. Relevant equations


    3. The attempt at a solution

    So I´m just thinking:

    an alpha-particle is two protons and two neutrons and the Si nucleus is 14 protons and 15 neutrons. Addition would give 14+2=16 protons and 15+2=17 neutrons.

    Hence, the result would be 33S. 33S is stable and naturally occuring (0.75%). Then one proton is emitted; subtraction one proton would give 32P. This is not naturally occuring and half-life of it is ~14 d. The rection would then be something like:

    29Si + [tex]\alpha[/tex] --> 33S* --> 32P + p

    Can one just add protons and neutronsl like this? The problem text states that the collision results in that a nuclear rection takes place. Can I assume that I am supposed to just add neutrons and protons like this? But why would the stable 33S emit a proton? And wouldnt it be strange to have the reaction product being the unstable 32P?

    Or could it possibly be that the collision results in a reaction where the only result is one proton being emitted, resulting in an Al nucleus?

    Either way, it seems TOO trivial and the hint with all the nuclei seems confusing.
    Do I need to somehow look closer into nuclear stability and decay series of sorts in order to solve this problem? Or should one, somehow, just realize that the reaction is something different than my idea of "simple addition" of alpha + Si and then subtraction of p? Other particles (beta, gamma)...are they not involved here at all?

    Maybe the reaction does not take the way over 33S? At first I wanted to call this 33S an excited sulfur nucleus, would that be correct terminology? I did add the "*" in the answer because of this thought...

    Help is appreciated! Thanks in advance!
    Last edited: Jul 9, 2010
  2. jcsd
  3. Jul 8, 2010 #2
    My best guess is that you're right; I'd have reasoned the same way. The (33)S is excited, or more likely never forms because the proton is knocked out immediately. But I have never taken a nuclear physics course.
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