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Hot cup of delicious Radium Water

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data
    The radium isotope [tex]^{223}{\rm Ra}[/tex], an alpha emitter, has a half-life of 11.43 days. You happen to have a 1.30 g cube of [tex]^{223}{\rm Ra}[/tex], so you decide to use it to boil water for tea. You fill a well-insulated container with 340 mL of water at [tex]16.0^\circ {\rm C}[/tex] and drop in the cube of radium.


    2. Relevant equations
    Alpha Decay:
    X (A, Z) yields Y (A-4, Z-2) + alpha + energy

    Energy = K_alpha = (M_x - M_y - M_He)c^2

    Nuclear Decay and Half-life
    N = N_0 e ^(-r/t) = N_0 e ^(-t/T) = N_0 (1/2)^(t)/(t/2)
    T = 1/r

    Heat Transformation?
    Q = E_th = McT

    3. The attempt at a solution
    I think is a heat transformation problem except I learned that 20+ chapters ago, last semester and need some advice.

    Since it's alpha decayer Radium (223, 88) becomes Rodium (219, 86) + alpha + energy. Loses 2 protons and 2 neutrons.

    E = (mass_Ra - mass_Rn - mass_He) * 931.5 MeV/u
    E = (223.018499 u - 219.009477 u - 4.002602) * 931.5 MeV/u
    E = 5.98 MeV

    Change in T = T = 84 Kelvin or Celsius

    Mass of Water = .998g/cm^3 * 340 mL * 1kg/1000g = .33592 kg

    Q = E_th = McT = (.33592 kg)(4190 J/kg K)(84 K) = 118230.4032 J

    Am I going about this the right way? This chapter is about radioactive decay and quantum physics. I can't seem to make the connection to alpha decay and heating water.
     
  2. jcsd
  3. Dec 9, 2007 #2

    mgb_phys

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    Yes, I think you can assume that all the energy from the alpha particle is absorbed by the water and so it is then a simple thermodynamic (Q = m C T) problem.
     
  4. Dec 10, 2007 #3

    Kurdt

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    If you're doing A2 physics there will be a synoptic element creeping in this year. This means you will use knowledge from a range of physics subjects to solve problems.
     
  5. Dec 10, 2007 #4
    This isn't working out. The energy output for the radium when calculated from MeV to Joules is
    5.98023 MeV * (1*10^6 ev / 1 MeV ) * (1.602 *10^-19 J / eV) = 9.58*10^-13 J

    The thermal energy required to heat the .335 kg of water in the problem is 1.18 * 10^5 J

    The energy output from the radium is once every half-life: 11.43 days or 987552 sec. At this rate the water won't heat in a reasonable amount of time.
     
  6. Dec 10, 2007 #5
    Is it? What's special about 1.30 g that means it gives you one decay every half life?
     
  7. Dec 10, 2007 #6

    dynamicsolo

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    I think an equation you haven't invoked yet which will be of help is the activity equation A = (lambda)N , where lambda is the (absolute magnitude) of the decay constant and N is the number of nuclei of the isotope in the sample. The activity A will be the number of decays per second (or Becquerels). You can find the decay constant from the half-life of the isotope. You know the mass of the sample and the molar mass of the radium isotope, so you can find N. You can thus find the number of decays per second and hence the radioactive decay power (the assumption is that the interval you are interested in is brief compared to the half-life, so the power is essentially constant). As pointed out, you are also assuming that all this power goes into heating the water.

    I believe 1.3 gm will more than suffice, though I'd personally refuse to drink the tea...
    (returns after computing the power output) Definitely use a set of tongs to handle the radium cube. (Also use the heavy ceramic mugs...)
     
    Last edited: Dec 10, 2007
  8. Dec 10, 2007 #7
    Thanks again but I think the units are confusing me here.

    lambda = r = 1 / T = ln 2 / t_(1/2) = 7.02*10^-7 [1/second]

    N = 1.30 g 223^Ra * 1 mole / g * atom / 6.022*10^23 mole = 9.68^-27 atom.
    1 atom has 1 nuclei ? so N is the same number nuclei as atoms?

    A = rN = 9.68*10^-27 nuclei decay /s ?

    A previous thing I did shows 9.58*10^-13 Joules per decay.

    So A * that = 6.51*10^-45 Joules / s

    Now according to Q_water = (.340 kg)(4190 J/kgK)(84K) = 118230 J

    so time = 118230 J / 6.51*10^-45 J/s = 1.82 *10^49 s

    still not right. are the units of N incorrect?
     
  9. Dec 10, 2007 #8

    dynamicsolo

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    Wouldn't you multiply the number of moles by Avogadro's number?

    N = (1.30 g 223^Ra / {223g/mole} ) x (6.022*10^23 nuclei/mole)
     
  10. Dec 10, 2007 #9

    Kurdt

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    If you have the energy required to heat the water which i think is correct, and you have how much energy each decay produces then you can work out how many atoms need to decay to heat the water. You can work out N0 from the mass of a single radium atom and the mass of the block of radium, then N is simply the energy needed divided by the energy per atom decay.

    You should be able to work out the rest. Activity isn't needed.
     
  11. Dec 10, 2007 #10
    to dynamic...
    I think your right. but even then with the new number I still get a wrong answer.

    to kurdt...
    I recalculated everything and stored it as exacts... maybe I didn't follow you correctly.
    Q = MCT = 119666.4 J {this is definitely correct}
    K_alpha_decay = 9.58*10^-13 J {also correct, triple checked}

    N = energy needed / energy per atom decay = Q / K

    N0 = mass_single radium atom / mass_given radium
    N0 = (223.02 u * 1.6605E-27 kg/u) / (1.30g * 1kg /1000g)
    N0 = 2.85 * 10^-22

    t(1/2) = 11.43 days = 987552 s

    Then from the book I have 2 choices of equations:
    N = N_0e^(-t/T)
    where T = time constant = t_(1/2) / (ln2)

    or less work...
    N = N_0 (1/2)^(t/(t/2))

    they both give the same answer though.

    t = [(t_1/2) ln(N/N_0)] / ln 2 = 126767742.9 seconds
     
  12. Dec 10, 2007 #11

    dynamicsolo

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    Shouldn't this be just the reciprocal of what you have?

    N0 = (0.0013 kg) / (3.70·10^-25 kg/atom)

    To Kurdt: Perhaps the activity isn't needed, but it gets to the answer pretty directly.
     
  13. Dec 11, 2007 #12
    To dynamic,

    By dimensional analysis so far:
    N = Q / K = [Joules]/[Joules/decay] = 1.25*10^17 [nuclie decay]
    N_0 = (.0013 kg) / (3.70*10^-25 kg/atom) = 3.51*10^21 [atom] = ?[nuclie decay]
    t_(1/2) = 987552 [seconds]

    N = N_0 (.5)^[t/(t_.5)]
    ln (N/N_0) = t/t_.5 ln (.5)
    t = [ln (N/N_0) / ln (.5) ] * t_.5
    t = 14595934.25 [seconds]

    still not right.
     
  14. Dec 11, 2007 #13

    dynamicsolo

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    It is exactly at this point that I proposed using the activity level of the sample in order to find the number of decays per second. Now that you have N_0, you can find the number of decays per second by multiplying that by the decay constant lambda. With the energy per decay, you can then find the rate of energy release (power) from the radium.

    How do this time have anything to do with the rate of heating of the water? You have to find the power being delivered from the radium sample. You know how much energy is required to heat the water to 100º C, so that energy divided by the heating power will give you the time sought in the problem.
     
  15. Dec 11, 2007 #14

    Kurdt

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    Thats my fault sorry. I said that N was the value I highlighted in red. Of course thats how many nuclei need to decay and thus N will be N0 - 1.25x1017.

    That gives a sensible answer.

    My apologies anyway.
     
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