Radioactive decay, need to find half life

[PhysicsMan999]

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.

 

[SammyS]

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 ?

 

[PhysicsMan999]

19.3 days

 

[SammyS]

Please explain what your variables represent.

Where did you get this equation?

R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)
​

 

[PhysicsMan999]

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.

 

[SammyS]

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.

 

[PhysicsMan999]

Okay, so I'm still not really sure how to solve it..plugging in the numbers just keeps on giving me 19.3..

 

[SammyS]

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.

 

[BvU]

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 ?