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Decay Energy/Gamma-ray production/Neutrino

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  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    dN/dt = -λN = -[itex]\frac{ln(2)}{τ(1/2)}[/itex]

    N(1 year) = N0e-λt

    Decay energy Q = KEf - KEi = mfc2 - mic2

    3. The attempt at a solution

    a): Using the above equations I found:

    56Co decays/s = 1.65 x 1047 decays/s
    56Ni decays/s = 1.20 x 1045 decays/s

    Q = c2(mf - mi)
    Since,
    mf = isotopic mass of 56Co = 55.940 u and
    mi = 55.942 u
    Q = 1.86 x 106

    I then use the equation, Radioactive power = W = Q[itex]\frac{A}{M}[/itex]
    Molar mass 56Co = 56 g/mol
    Molar mass 56Ni = 56 g/mol

    This is where I am stuck for part a). How do I implement the 0.75 M[itex]\odot[/itex] given in the question? And how do these units combine to give me power? Do I multiply the results I get with my W equation by 0.75 M[itex]\odot[/itex] to get eV/s? If so, then the rest of the problem should be simply converting units, which I can do easily.

    b): I'm having trouble determining the time they mean in this question. Do I use the .5 years, or do I have to calculate the rate at .5 years and use the resulting number? Additionally, as I am asking in another problem regaring neutrinos, how do I go about calculating the flux of the [itex]\gamma[/itex]-rays which could be observed on the earth?

    c): I believe this to be similar to how neutrinos are hard to observe. The energy of the [itex]\gamma[/itex]-rays is so great that it is hard to "capture" them to detect them. Like neutrinos, they must react with out atmosphere so we may detect their child particles. We detect less than the theoretical flux because we only have a limited number of detectors as well as our detectors do not detect each [itex]\gamma[/itex]-ray due to them not always reacting in our atmosphere (and above our detectors).

    Thank you for your help.
     
  2. jcsd
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