1. Limited time only! Sign up for a free 30min personal 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!

Radioactivity of Thorium question

  1. Aug 4, 2009 #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.)
    -- I got this correct (4.14 * 10^-5)

    (c) 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).
    What is the change in temperature of the water after 1 hr.?
    --------HELP: This is a multi-step problem. Among the things you must do are: i) determine the heat given off by the decay of one thorium atom, ii) determine the number of thorium atoms that decay to get the total heat given off, iii) determine the temperature change of the water knowing how much heat is given off.

    HELP: To determine the heat given off, figure out the difference in mass of the reactants and products and use E = mc2. Be careful of your units.

    HELP: To determine the amount of thorium that decays, figure out how many thorium atoms there were initially and then determine how many there are after 1 hrs. As always, watch your units.

    HELP: Once you know the heat released, use the heat capacity of water (4186 J/kg-K) and the mass of water to find the temperature change.

    2. Relevant equations
    N = NOe^-λt
    C= mcT
    E= mc2

    3. The attempt at a solution

    - 10g(1/228.028715)(6.022 * 10^23) = 2.64 * 10^22 atoms

    - N = NOe^-λt = (2.64*10^22)e^-(4.14 * 10^-5)(1) = 2.63989 * 10^22 atoms

    -- From this point I'm having problems figuring out the difference in mass between initial and final I think. Should the initial mass be: (2.64 * 10^22)(1 mo/ (6.022 * 10^23)(228.028715 g/ 1 mol) ...?

    What should the final mass be?

    I know from that point you plug in the change in mass into E = mc2, and take that energy and put it in for Q in Q=mcT (with the mass here being the mass of the water), and solve for T. The T I get here - 273 would be the degrees in celsius. Help!
  2. jcsd
  3. Aug 5, 2009 #2
    For one decay the difference in mass=mass of Th-(mass of Ra +mass of alpha)
  4. Aug 5, 2009 #3
    Ok. I ended up getting the right answer but don't fully understand the last step.

    mf= .005926u
    E=mc^2 = 8.9 * 10^-13 J

    For difference in # of atoms I did: Atoms initial = (10 g * 1 mol /232.038 g * 6.022 * 10^23) = 2.59526457 *10^22
    Atoms final= N = 2.59526457 * 10^22e^(-4.14 * 10^-5) = 2.59515713 * 10^22
    Atoms initial - Atoms final = 1.0744 * 10^18 atoms

    (Eatom)(difference in atoms) = (8.9 * 10^-13)(1.0744 * 10^18) = 956216 J

    Q= mcT
    956216 = (3.8)(4186)T
    T= 60.11

    ---- Why is it that I don't have to subtract 273 from this answer? Did I get my Q wrong, and got lucky with my answer? Since 4186 is in units J/Kg-K, it seems that the answer I'd get would be in K (hence subtract 273). Thanks for the help.
  5. Aug 5, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your expression is more correctly Q = mcΔT. You are calculating a temperature difference which makes the 273 drop out because it is added and subtracted.
  6. Aug 5, 2009 #5
    Thanks alot for the help. It makes alot more sense now.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Radioactivity of Thorium question
  1. Radioactivity question (Replies: 3)