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Homework Help: Radioactivity of Thorium

  1. Jul 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.028715 u) undergoes alpha decay and produces radium (atomic mass 224.020186 u) as a daughter nucleus. (Assume the alpha particle has atomic mass 4.002603 u.)

    (a) What is the decay constant of thorium? (Note that the answer must be in units of hrs-1.)
    Answer: 4.136E-5 hrs-1

    The energy released from the decay of 10 g of thorium is used to heat 3.8 kg of water (assume all the energy released goes into the water).

    (c) What is the change in temperature of the water after 1 hr.?


    2. Relevant equations
    E=mc2
    N0= Mtotal/Mparticle
    N=N0e^-λt
    Q=mcT

    3. The attempt at a solution
    N0= Mtotal/Mparticle
    N0=[(0.01kg)]/[(228.0287u)(1.66E-27kg)]
    N0=2.64E22 atoms

    N=N0e^-λt
    N=(2.64E22 atoms)e^-(4.136E-5 hrs-1)(1hr)
    N=2.642E22 atoms

    mi=228.028715u

    mf=224.020186u + 4.002603u
    mf=228.022789u

    m=mf-mi
    m=228.022789u - 228.028715u
    m=0.005926u

    E=mc2
    E=(0.005926u)[(931.5Mev/c2)/u]c2
    E=5.52Mev

    (5.52Mev)(1.602E-13J/Mev) = 8.843E-13 J

    (8.843E-13J)(2.642E22 atoms)=2.336E10J

    Q=mcT
    2.336E10J=(3.8kg)(4186 J/kg-K)(T)
    T=1468755.878 K = 1468482.878 C

    Can someone help me catch my mistake? I'm not sure what I did wrong. Thank you.
     
  2. jcsd
  3. Jul 28, 2008 #2

    Astronuc

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    Staff Emeritus
    Science Advisor

    This assumes all of the Th-228 decays.

    One starts with No = 2.642E22 atoms, but how many atoms remain at the end of 1 hr? The difference No - N (1 hr) = the number of atoms that have decayed.

    It's not that 10 g of Th-228 decayed, but a certain fraction of atoms in Th-228 have decayed, and that is related to the activity.
     
  4. Jul 28, 2008 #3
    Thank you. I got it now.
     
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