1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Decay - How does this notation represent a reaction?

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\mbox{What is the atomic number Z, the atomic mass number A, and the element X in the reaction} \ _{5}^{10}B(\alpha,p)_{Z}^{A}X[/tex]

    2. Relevant equations

    If my understanding of the question is correct, then we are dealing with alpha decay and beta-plus decay.

    3. The attempt at a solution

    I can't find an example of this notation for the reaction in my textbook, so I assumed that it meant:

    [tex]_{5}^{10}B \ \mbox{undergoes alpha decay and the resultant product undergoes beta-plus decay to form} \ _{Z}^{A}X[/tex]

    And got the following:

    [tex]_{5}^{10}B \ \mbox{undergoes alpha decay}[/tex]

    [tex]_{5}^{10}B \rightarrow _{3}^{8}Li + ^{4}He[/tex]

    [tex]\mbox{and} \ _{3}^{8}Li \ \mbox{undergoes beta-plus decay}[/tex]

    [tex]p \rightarrow n + e^+ + v[/tex]

    [tex]\mbox{So,} \ _{3}^{8}Li \ \mbox{transforms as follows}[/tex]

    [tex]_{3}^{8}Li \rightarrow _{2}^{8}He + e^+ + v[/tex]

    [tex]\mbox{Finally,} \ A = 8, Z = 2, X = He.[/tex]

    Please tell me if my interpretation of the notation (and subsequently my reasoning) is indeed correct.

    Thanks
    phyz
     
  2. jcsd
  3. Jul 7, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    This notation is peculiar to nuclear astrophysicists, I think; I'm not sure other nuclear physicists notate reactions quite that way. (It's handy, though, when you have to write down the various chains of reactions that take place in stars, since some of the reaction networks can run to dozens of processes...)

    So most people would write [tex]
    \ _{5}^{10}B + \alpha \rightarrow p + _{Z}^{A}X[/tex]. There are 14 nucleons involved here, 7 of them being protons, so the product should have 13 nucleons, of which 6 are protons.

    Your first step should be a combination of boron-10 with an alpha particle, to produce nitrogen-14:

    [tex]
    _{5}^{10}B + _{2}^{4}He \rightarrow _{7}^{14}N [/tex] .

    Apparently, the nitrogen-14 is "excited" and releases a proton, leading to

    [tex] _{7}^{14}N \rightarrow _{1}^{1}p + _{6}^{13}C [/tex] .

    Since the nitrogen nucleus doesn't exist for long, it is simply omitted here and the net reaction alone is presented.
     
  4. Jul 7, 2008 #3
    Thank you very much for an excellent explanation.

    (Turns out it's a good thing I asked hey? :wink:)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?