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Homework Help: Nuclear effect from 0.25 µg 220-Rn

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm a bit negative to this assignment, because I think this has lack of information. The question is:

    "220-Rn is radioactive and gives α-particles with the energy of 6.4 MeV. An injection of 0.25 µg Radon is performed in a closed container. How big effect is developed in the container immediately after the injection because of the α-radiation?

    2. Relevant equations

    220-Rn --> 220.0114 u --> Half life: 56 s

    3. The attempt at a solution

    First I must determine the number of nucleus: 0.25 µg is 2.5*10^-16 kg. And the weight of one nucleus in kg is 220.0114*1.6605402*10^-27 = 3.65337774*10^-25 kg.

    Now, my weight of used nucleus divided by the weight of one:

    (2.5*10^-16)/(3.65337774*10^-25) = 684298251.6 pieces of 220-Rn.

    I was trying to multiply the number of nucleus by 931.49 MeV, but I know this was wrong.

    What I'm missing to be able to answer in Watts is over which time this energy was obtained. Anyone who agree?

    Maybe someone know how to solve this probably easy problem?
  2. jcsd
  3. Jun 9, 2013 #2


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    How many grams are in 1 microgram? How many grams are in 1 kg? Check your unit conversions.

    Remember that half of the nuclei have decayed after the half life period has elapsed.
  4. Jun 9, 2013 #3


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    Suppose 220-Rn decays at rate λ, i.e. in a short period of time δt, Nλδt atoms of a large sample of N will decay. What fraction will have decayed after an extended time t? What relationship does that give you between λ and the half life?
  5. Jun 10, 2013 #4
    0.25 µg is 2.5*10^-7 --> 2.5*10^-10 kg.

    First, I was counting /1000 for every step from 2.5*10^-7, but there's only one /1000 step.
  6. Jun 10, 2013 #5
    Of course there's 6.842982516*10^14 pieces of 220-Rn.
  7. Jun 11, 2013 #6
    I suppose this is the solution:

    The half life of 220-Rn is 56 seconds. In these 56 seconds, half of the nucleus have decomposed, and I have 6.842982513*10^14 pieces of them. Half of these is 3.421491257*10^14 pieces. I multiply these by 6.4 MeV, which is 2.189754404*10^15 MeV. This is the energy developed in 56 seconds in MeV, and Watts are Joules per second. First, the energy in Joules is:

    (2.189754404*10^15)*(1.602*10^-13) = 350.7986555 Joules

    This value is over 56 seconds, and per second this energy is:

    6.264261706 Watts.

    Can someone please check that this is correct?
  8. Jun 11, 2013 #7


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    No, you can't turn a half life into an instantaneous rate quite that easily. See my previous post (#3). E.g. the rate will be twice as much at the start of the 56 seconds as it will be at the end. You might have been given a formula for making this conversion, or you can figure it out with calculus.
  9. Jun 12, 2013 #8
    Ok, I was thinking that the decay was emitting this energy. I have this initial energy without decay (it starts decay immediately, I know), and this was actually my first own theory. My second own theory was that the energy comes out from the decay, but it's not.

    If the question is based on Watts, I must know the time. At least I have 6.842982513*10^14 inital nucleus, which each gives 6.4 MeV. Totally, they are giving 4.739508808*10^15 MeV, which in Joules is: 701.5973111 J.

    If this is for 56 seconds, this would give 12.52852341 Watts.
  10. Jun 12, 2013 #9


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    No, you've missed the point of my comment completely. The decay is emitting the energy, of course. The point is that you are not being asked for the average rate over 56 seconds. You are being asked for the rate right at the start, and that's double what it will be after 56 seconds, so must also be a lot more than the average over 56 seconds.
    If you don't have a formula for converting between half life and instantaneous decay rate you'll need to work it out using differential equations. It's not hard.
    Suppose the remaining quantity is N = N(t). In a small time interval δt a fraction λ decays. So at time t+δt, Nλδt have decayed:
    N(t+δt) = N(t)(1-λδt)
    Can you turn that into a differential equation and solve it?
  11. Jun 13, 2013 #10
    Thanks. I knew this was the average value, and I was thinking how to give this at a specific second. I checked my old mathematic litterature, in which I last did differential equations in 2006. I'm not so good as I was.

    To put it as simply as I can, I would do it like this:

    N = 6.842982513*10^14 nucleus


    Within the first second, it must be:

    N*0.5^(1/56) = 6.758804655*10^14 nucleus

    Then, it must be the initial amount of nucleus minus this result:

    (6.842982513*10^14) - (6.758804655*10^14) = 8.4177858*10^12

    This number of nucleus is decaying within the first second.

    The energy emitted the first second is then: (8.4177858*10^12)*6.4 = 5.387382912*10^13 MeV

    This in Joules is: 8.630587425 Joules.

    Because this is within a second, the value in Watts is 8.630587425 Watts.
  12. Jun 13, 2013 #11


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    That's much better, and quite likely good enough. But it is not necessary to pick an arbitrary period such as 1 second. Why not 0.1 second, etc?
    If the instantaneous decay rate is λ then after time t the fraction e-λt remains. If the half life is h we have e-λh = 0.5, so λ = ln(2)/h. (In your first attempt you were effectively using λ = 1/2h.)
  13. Jun 14, 2013 #12
    Thanks a lot for the help. I was also thinking that a good way to really check this value, is to do following; after 56 s the decay is half of the decay at the start:

    N*0.5^(56/56) - N*0.5^(57/56) = 0.0061506702 N

    N*0.5^(0/56) - N*0.5^(1/56) = 0.0123013405 N

    The first value is exactly half of the second value. Beautiful.
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