Certain isotope tell you that the decay rate decreases

  • Thread starter xinlan
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  • #1
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Homework Statement



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.

Homework Equations



R=Ro.e^-lamda/t
then T1/2 = ln(2)/lamda

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:
R=Ro.e^-lamda/t
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..
 
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Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
19,274
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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.
 
  • #3
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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?
 
  • #4
61
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I got it.. :)
 

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