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Certain isotope tell you that the decay rate decreases

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Measurements of a certain isotope tell you that the decay rate decreases from 8337 decays/minute to 3174 decays/minute over a period of 5.00 days.

    2. Relevant equations

    then T1/2 = ln(2)/lamda

    3. The attempt at a solution
    I converted all the units..
    t = 5 days = 432000 seconds
    R = 8337 decays/minute = 52.9 decays/s
    Ro=3174 decays/minute = 138.95 decays/s
    I did:
    52.9=138.95 . e^-lamda*432000
    lamda = -1.86*10^-6

    I am confused.. the lamda should not be a negative number right?

    because half-life should be a positive number..
    please help me..
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Apr 14, 2008 #2


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    Staff: Mentor

    Well, if one take R0 to be the lesser activity and the later activity to be greater, then one will get a negative lamdba since the activity is increasing.

    Usually R(t) < R0, i.e. R(t) is decreasing from R0.

    If one takes the ln of the decay equation, one can make it linear.

    Start with R0 = 8337 decays/minute, then R(t=5days) = 3174 decays/minute, but make sure units/time are on the same basis.
  4. Apr 14, 2008 #3
    I should convert all the units right?

    then if the lamda is negative, the half-life will be negative, and should I just plug in the minus sign as well on the computer?
  5. Apr 15, 2008 #4
    I got it.. :)
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