Radioactivity and decay constant

1. May 26, 2009

sweminka

A piece of radioactive substance gives a received count rate of 6000 counts per minute in a detector whose efficiency is known to be 5%. If the sample contains 10^10 atoms, what is the decay constant ( λ ) of this radioactive substance ?

No idea how to solve this problem. Any help would be really appreciated.

Edit: I am a medical student studying abroad, and the physics teacher doesn't speak English properly. Please believe that I have tried to understand, but after 1 day of looking for information on the internet, I still have no idea how to solve this problem.

Last edited: May 26, 2009
2. May 26, 2009

fatra2

Hi there,

To answer your question, there are many assumptions to be made:
1. I assume 6000 counts/minute in the first minute.
2. $$10^{10}$$ atoms is supposed to be be the initial excited/radioactive atoms in the sample
3. 6000 counts/minute assumes that all radiation emitted passes through the detector, with a 5% efficiency.

If all these assumptions are true, then you can simply apply the radioactive decay equation to your problem and the answer is solved in two lines.

3. May 26, 2009

sweminka

Here are my calculations :

N(t)=No*exp(-λt)

10^10 - 6000 = (10^10) * exp (-λ*60)

( 10^10 -6000) / (10^10) = exp (-λ*60)

ln ( ( 10^10 -6000) / (10^10) ) = -λ*60

λ= ( - ln ( ( 10^10 -6000) / (10^10) ) ) / 60

λ = 10*(-8) per second

Last edited: May 26, 2009
4. May 26, 2009

sweminka

The answer is supposed to be 2*10^(-7) though :(

I guess it has something to do with the detector's efficiency ?

Last edited: May 26, 2009
5. May 26, 2009

fatra2

Hi there,

You are forgetting the efficiency of the detector in your calculations. Don't forget that only 5% of the particles are detected.

6. May 26, 2009

sweminka

"only 5% of the particles are detected" This sentence suddenly made everything clear to me. I guess I couldn't understand the meaning of "efficiency". Anyway, I replaced 6000 by 120 000 (100%) in my calculations, and I find the correct answer.

Thank you very much :)