Radioactive decay, need to find half life

AI Thread Summary
The discussion revolves around calculating the half-life of a radioactive material that decreases to one-fifteenth of its original amount in 26 days. Participants clarify the use of the decay equation and the variables involved, including the decay constant and the relationship between decay rate and half-life. One participant suggests that the calculated half-life of 19.3 days is incorrect, proposing instead that it should be around 6.5 days based on estimations of decay over multiple half-lives. The correct approach involves solving the equation for the decay constant and then determining the half-life from it. The conversation emphasizes the importance of understanding the underlying equations and concepts in radioactive decay.
PhysicsMan999
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Homework Statement



  1. The number of radioactive nuclei in a particular sample decreases over a period of 26 days to one-fifteenth of the original number. What is the half-life of the radioactive material, in days?

Homework Equations


R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)

The Attempt at a Solution


DN= N - (1/15)N
DT=26 days
Plugged these in and solved for T(1/2), but CAPA says my answer is wrong.
 
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PhysicsMan999 said:

Homework Statement



  1. The number of radioactive nuclei in a particular sample decreases over a period of 26 days to one-fifteenth of the original number. What is the half-life of the radioactive material, in days?

Homework Equations


R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)

The Attempt at a Solution


DN= N - (1/15)N
DT=26 days
Plugged these in and solved for T(1/2), but CAPA says my answer is wrong.
What did you get for half-life ?
 
19.3 days
 
Please explain what your variables represent.

Where did you get this equation?
R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)
 
R=rate of decay
DN= change in number of nuclei
DN=change in time
lambda=decay constant
N=number of nuclei
T1/2= half life
I got the equation from a slide my prof has for our lectures.
 
PhysicsMan999 said:
R=rate of decay
DN= change in number of nuclei
DN=change in time
lambda=decay constant
N=number of nuclei
T1/2= half life
I got the equation from a slide my prof has for our lectures.
Starting with the following (which is equivalent to ##\displaystyle\ N=N_0\,e^{-\ln(2)\,t/t_{1/2}}## )
##\displaystyle\ N=N_0\, e^{-\lambda t}\ ##, where N0 is the number of nuclei at time, t = 0 , (the start),​
and taking the derivative w.r.t. time, we get:
##\displaystyle\ \frac{dN}{dt}=-\lambda N_0e^{-\lambda t}\ =-\lambda N\ .##
Also, ##\displaystyle\ -\lambda N=-\ln(2)/t_{1/2}\ .##​
This is the decay rate, similar to yours, except for a sign. However, it's the instantaneous decay rate, not the same as ΔN/Δt .

By the way, 19.3 days is way off for the half-life. At that rate it would take 38.6 days (2 half-lives) to get to 1/4 the original number of nuclei.

Getting to 1/16 in 28 days would give a half-life of 7 days.
 
Okay, so I'm still not really sure how to solve it..plugging in the numbers just keeps on giving me 19.3..
 
PhysicsMan999 said:
Okay, so I'm still not really sure how to solve it..plugging in the numbers just keeps on giving me 19.3..
Solve ## \displaystyle\ N=N_0\, e^{-\lambda t}\ ## for λ , if t = 26, and N = N0/15 .

Then use your equation to find the corresponding half-life.
 
Dear PM,

In physics it's always good to make an estimate.
You know that after one half-life time, the activity (hence also the number of active nuclei) drops to one half the original number.
After two to one fourth
After three to one eighth
And after four to one sixteenth. So your answer should be close to 26 days / 4 = 6.5 days, and a bit lower.

From the above: Activity = $$A(t) = A(0) / 2^{\;(t/\tau_{1/2})}$$ or$$A(0)/A(t) = 2^{\;(t/\tau_{1/2})}$$ so what you want to solve is ## 2^{\;(26 {\rm \; days}/\tau_{1/2})} = 15## . That sound good ?
 
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