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Radioactive decays - given power value

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The masses of the following atoms are:

    238 U → 238.050786(2)
    234 Th→ 234.043598(5)
    4 He → 4.002603

    a) Calculate the energy available in the decay of 238U.

    b) Calculate the decay rate of 238U

    238U half time = 4.468*10^9yrs

    c) If a block of uranium produces 1mW of power, how many disintegrations per second is this and what is the activity in Curies. Neglect other decays other than the one above.

    d) Calculate the number of uranium atoms and the mass of the block.

    - This is a question for an upper level medical physics class, but it seems to be relatively easy. So I am not sure if I am putting this in the right place.

    2. Relevant equations
    Mass calculation of radioactive decay

    1 amu = 1.66 * 10^ -27 [kg]

    t½ = ln2 / λ

    A = λ*N = ΔN/ Δt

    1 Ci = 3.7* 10^10 [s^-1]

    3. The attempt at a solution

    a)
    Δ amu = 238.050786(2) - 234.043598(5) - 4.002603 [amu] = 4.58*10^-3 [amu]

    Δ mass = Δ amu * 1. 66*10^-27 [kg]
    Δ mass = 4.58*10^-3 * 1.66 *10^-27 = 7.6106 *10^-30 [kg]

    E = Δ mass * c^2
    E = 7.6106 *10^-30 * 3*10^8 = 2.28 * 10^-21 J

    b)

    Δ t = 4.468*10^9 * 365 * 24 *60 * 60 = 1.409028* 10^17
    Δ N = 6.022*10^23 atoms/ Δ amu = 7.9126* 10^-8

    decay rate = ΔN/ Δt = 5.615 * 10^ -25

    c)

    disintegration per second = A = λN and 1 Ci = 3.7 * 10^10 [s^-1]

    But how does power play a role? Does power affect the number of atoms? If yes how?
    Assuming power has nothing to do with N, I am approaching the problem like this:

    t½ = ln2/ λ
    λ = 4.91932 * 10^ -18 [s^-1]
    N = 6.022 * 10^ 23/ 238

    A = λN = 12447 Bq = 3.36 * 10^ -7 [Ci]

    d) since I don't know how power is related, I don't really know :(
     
  2. jcsd
  3. Feb 21, 2016 #2

    mfb

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    Staff: Mentor

    Where did the square of the speed of light go?
    What did you calculate here? Working with units would make it easier to follow.
    That is not correct. You can directly get the decay rate from the half-life without other quantities involved.
    If every apple has a mass of 0.2 kilogram and I give you 5 kg per second ("power"), how many apples per second do you get?
    Power does not "affect" the number of atoms but you need power to figure out how many decays happen per second.
     
  4. Feb 21, 2016 #3
    Thank you for correcting me my mistakes, and helping me to move the thread to the right place!

    E = Δ mass * c^2
    E = 7.6106 *10^-30 *( 3*10^8)^2 = 6.84*10^-13 J

    Δ N = 6.022*10^23 [atoms/mol] / Δ amu [g/mol] = 7.9126* 10^-8 [atoms/g]

    Regarding the decay rate, I thought it means activity, which is A = λN. How can I get directly from half life? Can you give me some pointers?
    t½ = ln2/ λ ==> λ = 4.91932 * 10^ -18 [s^-1]

    For power, does that mean for question c)
    Disintegration per second = A = 1*10^-3 [s^-1] ?

    Thank you so much!
     
  5. Feb 22, 2016 #4

    mfb

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    That would give 12 tons per atom (the inverse value). Very heavy atoms...

    Decay rate: Hmm, could be a matter of definition. If you used it as decays per second per gram, then you need the number of atoms per gram, sure.
    No. Where does that value come from?
     
  6. Feb 22, 2016 #5
    Oh I see that doesn't make sense now...

    So am I correct to think that for decay rate A = λN
    t½ = ln2/ λ => λ = 4.91932 * 10^ -18 [s^-1]
    Δ N = 6.022*10^23 [atoms/mol] / 238 [g / mol] = 2.530*10^21 [atoms/g]

    A = 12447 [s^-1] = 1.24*10^4 [s^-1]

    That value comes from the power value 1 mW = 1*10^-3 W

    This question seems really easy, but I haven't done physics in a while and couldn't really wrap my head around.
    Thanks so much for your help!
     
  7. Feb 22, 2016 #6

    mfb

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    Another value that comes out of the blue...
    W is not the same as 1/s.

    All the quantities you need can be derived by combining given values in a way that the units match. If the units do not match, the answer is certainly wrong.
     
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