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Radioactive decay

  1. Dec 23, 2011 #1
    d1. The problem statement, all variables and given/known data
    the half life of U234 is 245.5 years. It decays by α emmission to Th230 whos half life is 75.38. there are 10^6 atoms of U234 and 0 atoms of Th230 at t=0.
    At what time will the number of atoms of each type be equal?

    the half life of U234 is 245.5 years and half life of Th230 is 75.38
    λa for U is 2.82e^-3 and λb for Th is 9.2e^-3


    2. Relevant equations
    Na = No*exp(-λut)
    Nb = No*(λa/(λb-λa))*(exp(-λa*t)-exp(-λb*t))



    3. The attempt at a solution
    They Na and Nb are equal when Na = Nb

    therefore:
    No*exp(-λat) = No*(λa/(λb-λa))*(exp(-λa*t)-exp(-λb*t))
    Divide by No and exp(-λat)
    1 = (λa/(λb-λa))*(1-exp-(λa + λb)t)
    Divide by (λb-λa)/a

    1 - ((λb-λa)/λa) = exp-(λa + λb)t
    getting t on its own
    t = ln(1-((λb-λa)/λa)) / -(λb + λb)

    t = ln(1-((9.2e-3-2.82e-3)/2.82e-3)) / -(2.82e-3 + 9.2e-3)
    t = ln(1-(0.00638/2.82e-3)) / (- 0.01202)
    t = ln(1-2.26) / (-0.01202)
    t = ln(-1.26)/-.01202

    here is where the problem is as you can not get the ln of a negitave number
    Can anyone figure out where i have gone wrong?
    or is there a problem with the question itself
     
    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 23, 2011 #2
    Hi Howlin!!

    Your solution was a bit messy for me to understand :tongue2: so i couldn't figure out your error ...
    however i can tell you how i would have solved it ,,,

    You can write rate law of decay of U234 as
    [itex]N = N_O e^{-\lambda_{U} t}[/itex]

    here N is number of atoms of U left

    at the time when both U and Th have same number of atoms [itex]N\frac{N_O}{2}[/itex] as half of U would give same number of Th atoms

    now you can solve for t ...
     
  4. Dec 23, 2011 #3
    If you go to
    http://www-naweb.iaea.org/napc/ih/documents/global_cycle/vol I/cht_i_06.pdf

    equation 6.5 and 6.18 are the two i used with the exception for 6.18 that i did not include the +etc in it

    When the amount of the two materials are the same Na and Nb are equal the two equations are equal

    so 6.5 = 6.18
    you can rearrange the equations to get t on its own so you can find out when the time when the two substances have the same amount

    (you still with me, what i done was just work out how i got to the final equation)

    its just that when i put in my values for λ, the equation does not work
    has this helped at all?
     
  5. Dec 23, 2011 #4

    Redbelly98

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    That won't work here. At some time, half the U atoms are left. Question #1: Does that mean the remaining half of the original atoms are all Th? Answer: no. Question #2: Why not?

    I recommend writing two differential equations, one for the number of Uranium atoms and another for the number of Thorium atoms:

    [tex]\frac{dN_U}{dt} \ = \ ???[/tex]
    [tex]\frac{dN_{Th}}{dt} \ = \ ???[/tex]
     
  6. Dec 24, 2011 #5

    Curious3141

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  7. Dec 24, 2011 #6
    your Nb equation is one of the equations i am using and the other is the Na = Noexp(-λt)

    We know that the amount of substance a remaining after time t is Na and the amount of substance b that is left after time t is Nb but i have to figure out after how long will both substances have the same number of particles?

    This occurs when Na = Nb
    I then made the two equtions equal to find out at what time this will occur
    but when i rearrange the equations to get t on its own and work it out to get a value for t
    this t value should be the time when both substances have the same atoms but i can not get this to work out for some reason and want to figure out why

    curious 3141 can you work out an equation to find out at what time will two substances (A and B) will have the same amount of atoms if substance A decays into substance B and substance B decays into substance C

    I have my answer in the first post and want to see if you get the same equation
     
  8. Dec 24, 2011 #7

    Curious3141

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    OK, I see that. Your working is too messy to easily figure out and there are some obvious typos. Please format it into LaTex if you want us to review it.

    That having been said, I believe the solution here is that there will never be a time when there are equal quantities of the two elements. For equality to exist, this condition must be met:

    [tex]\lambda_A > \frac{1}{2}\lambda_B[/tex]

    or, equivalently, in terms of the respective half-lives:

    [tex]T_A < 2 T_B[/tex]

    This condition is not met here. You will not get a real solution for the time.

    Provided the condition is met, the time you want is [itex]\tau[/itex] where:

    [tex]\tau = \frac{1}{\lambda_A - \lambda_B}\ln(\frac{2\lambda_A - \lambda_B}{\lambda_A})[/tex]
     
    Last edited: Dec 24, 2011
  9. Dec 24, 2011 #8
    At what time will hte number of atoms of both substances be equal if at t= there are 10^6 atoms of substance A and 0 atoms of substance B and substance A has a half life of 245.5 years and substance B has a half life of 75.38 years.

    Here are the two equations for determining how many atoms of each substance is left at a particlar time


    [itex]N_{A}[/itex] = [itex]N_{O}[/itex][itex]e^{-λ_{A}t}[/itex]

    [itex]N_{B}[/itex] = [itex]N_{O}[/itex][itex]\frac{λ_{A}}{λ_{A}-λ_{B}}[/itex]([itex]e^{-λ_{A}t}[/itex] - [itex]e^{-λ_{B}t}[/itex])

    The time when both substances are equal is when

    [itex]N_{A}[/itex] = [itex]N_{B}[/itex]

    Therefore

    [itex]N_{O}[/itex][itex]e^{-λ_{A}t}[/itex] = [itex]N_{O}[/itex][itex]\frac{λ_{A}}{λ_{A}-λ_{B}}[/itex]([itex]e^{-λ_{A}t}[/itex] - [itex]e^{-λ_{B}t}[/itex])

    Since [itex]N_{O}[/itex] and [itex]e^{-λ_{A}t}[/itex] are on both sides then we can gt rid of them and the equation becomes

    1 = [itex]\frac{λ_{A}}{λ_{A}-λ_{B}}[/itex](1 - [itex]e^{-(λ_{A} +λ_{B})t}[/itex])

    We rearrange the equation to get the t value on its own on one side and the rest of the values on the other side

    We divide by [itex]\frac{λ_{A}-λ_{B}}{λ_{A}}[/itex]

    1 - [itex]\frac{λ_{A}-λ_{B}}{λ_{A}}[/itex]= [itex]e^{-(λ_{A} +λ_{B})t}[/itex]
    We get rid of the exp

    ln(1 - [itex]\frac{λ_{A}-λ_{B}}{λ_{A}}[/itex]) = [itex]-(λ_{A} +λ_{B})t[/itex]

    getting t on its own

    t = [ln(1 - [itex]\frac{λ_{A}-λ_{B}}{λ_{A}}[/itex]) ] / [itex]-(λ_{A} +λ_{B})[/itex]

    now the value we get for t should be the time when both substances have the same amount of atoms

    But when i try and use this equation for two differnt questions (the one above and a different one where i know the answer) the equation do not give me the correct t value
     
  10. Dec 24, 2011 #9

    Curious3141

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    It's almost two in the AM here, so I can only take a cursory look at this. But right off the bat, there are two glaring errors:

    The equation for [itex]N_B[/itex] should read: [itex]N_{B}[/itex] = [itex]N_{O}[/itex][itex]\frac{λ_{A}}{λ_{B}-λ_{A}}[/itex]([itex]e^{-λ_{A}t}[/itex] - [itex]e^{-λ_{B}t}[/itex])

    Note the transposition of the lambdas. The equation you quoted in your first post was right (which also matches up with the equation I derived in my thread). You can see immediately (I hope) why this has to be the right form. If lambda A is higher than lambda B, the difference between the exponential terms would become negative. But the denominator of lambda B minus lambda A would also be negative, giving an overall positive value for N(B). Everything also works out to be positive in the other case if lambda B were to be higher than lambda A, so this is the correct form. In the version you started with, N(B) is always negative!

    If you started wrong, the rest is going to be wrong. Nevertheless, here's another error, algebraic this time:

    When you divide [itex]N_{O}[/itex][itex]e^{-λ_{A}t}[/itex] = [itex]N_{O}[/itex][itex]\frac{λ_{A}}{λ_{A}-λ_{B}}[/itex]([itex]e^{-λ_{A}t}[/itex] - [itex]e^{-λ_{B}t}[/itex])

    by [itex]N_{O}[/itex] and [itex]e^{-λ_{A}t}[/itex],

    your result should be: 1 = [itex]\frac{λ_{A}}{λ_{A}-λ_{B}}[/itex](1 - [itex]e^{(λ_{A} -λ_{B})t}[/itex])

    See if you can see why. Here, I left the lambda terms reversed erroneously so as not to complicate your understanding, but when you redo it, make sure to start with the right equation.

    There may be other errors, but this should be enough to start you off. If you're having trouble with the algebra, I suggest you be careful and work one step at a time.
     
    Last edited: Dec 24, 2011
  11. Dec 24, 2011 #10
    Im sorry when i was typing the equation I made the mistake and the Nb one that u have is the one i have been using

    I just noticed that when dividing by exp(-λt)
    I forgot about the a negitave sign when dividing

    thanks for all your help
    It was a slide from a lecture and was basicly copying him but then when i went step by step he made a mistake when dividing by [itex]e^{-λ_{A}t}[/itex]
    Thank you for all your help and happy holidays
     
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