How to Determine the Decay Constant of a Radioactive Substance?

AI Thread Summary
To determine the decay constant (λ) of a radioactive substance, one must consider the efficiency of the detector, which in this case is 5%. The initial count rate of 6000 counts per minute needs to be adjusted to reflect the total emitted radiation, resulting in an effective count of 120,000 counts per minute. By applying the radioactive decay equation, the correct decay constant can be calculated as λ = 2 x 10^(-7) per second. Understanding the detector's efficiency was crucial in solving the problem accurately. This highlights the importance of accounting for all variables in radioactive decay calculations.
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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.
 
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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.
 
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
 
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The answer is supposed to be 2*10^(-7) though :(

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

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