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## Homework Statement

" Independent measurements are taken on two identical radioactive samples. On the first radioactive sample, a radiation detection system measures

*N*decays on a radioactive sample during a total period of

*T*. On the second radioactive sample, a radiation detection system measures 2

*N*decays on a radioactive sample during a total period of 3

*T*. What is the half-life of the radioactive sample? "

## Homework Equations

Number of particles decayed ## = \Delta N = N_0 (1-e^{-\lambda t}) ##

Half life = ## \frac{ln2}{\lambda} ##

decay constant = ## \lambda ##

## The Attempt at a Solution

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My reasoning is, since the question states that the sources are identical, this means that the isotopes are identical. I assume the measurements are begun at the same time i.e. ## T_0 = 0## for both of them. The number of particles cannot then be the same because the counts per minute for source 1 is ## cpm_1 = \frac{N}{T} ## and the counts per minute for source 2 is ## cpm_2 = \frac{2}{3} \frac{N}{T} = \frac{2}{3} cpm_1##, and therefore the sources must have different activities since they are otherwise identical, and therefore the only thing that can be causing the change in count rate is that there is a different number of particles ## N_0 ## in each source.

My attempt at the solution was this: the number of decays in each source is proportional to ## 4\pi r^2 ## times the counts detected. ## k ## is the proportionality constant which divides out.

# of decays in source 1:

$$N = k{N_0}_1 (1- e^{- \lambda T})$$

# of decays in source 2:

$$2N = k{N_0}_2 (1- e^{- \lambda (3T)})$$

Plugging the equation for ## N ## into the second equation:

$$2k{N_0}_1 (1-e^{-\lambda T}) = k{N_0}_2 (1-e^{-\lambda 3T})$$

rearranging for ##e^{- \lambda T}## I get a cubic equation:

$$(e^{-\lambda T})^3 - 2\frac{{N_0}_1}{{N_0}_2} e^{-\lambda T} + (2\frac{{N_0}_1}{{N_0}_2} -1) = 0$$

I guess if I had numbers i could plug this into a calculator to get the zeros but I'm not really sure if this is the right thing to do anyways.

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