# Age of radioactive sample from sample

#### Incand

1. Homework Statement
About a day after the Chernobyl disaster the radioactive fallout landed in Sweden. The relative activity of $^{133}I (t_{1/2} = 20.8h )$ and $^{131}I (t_{1/2} = 8.02 d)$ were measured at 28/4 17:00 at $270$ mBq and $1000$ mBq respectively.
Determine the time of the reactor failure. The exchange for creating different isobars with fission using thermal neutrons of $^{235}U$ can be found in the chart of nuclides.

(I'm afraid I made a rather poor translation but I'm sure the meaning comes across)

2. Homework Equations
$N(t) = N_0e^{-\lambda t}$

Half-life
$t_{1/2} =\ln (2) / \lambda$

Nuclear activity
$A(t) = \lambda N(t)$
3. The Attempt at a Solution
Lets refer to $^{133}I$ as element 1 and $^{131}I$ as element 2.
The ratio of the activities is by the law of radioactive decay is then
$R = \frac{\lambda_1 N_1(t)}{\lambda_2 N_2(t)} = \frac{\lambda_1 N_1(0)}{\lambda_2 N_2(0)} e^{t (\lambda_2-\lambda_1)}$
Rearranging we have
$t = \frac{1}{\lambda_2 - \lambda_1}\ln \left( R\frac{\lambda_2 N_2(0)}{\lambda_1 N_1(0)} \right)$

We know that $R=0.27$ and the decay constants can be computed too
$\lambda_1 = \log (2) /(20.8 \cdot 3600) \approx 9.2568\cdot 10^{-6}$
$\lambda_2 = \log(2)/(8.02 \cdot 3600 \cdot 24) \approx 1.0003\cdot 10^{-6}$.

To find the initial ratio of the elements I looked at the ratio in the Thermal fission yield column here https://www-nds.iaea.org/sgnucdat/c3.htm
$\frac{N_2(0)}{N_1(0)} \approx \frac{2.878}{6.59} \approx 0.4367$.

Inserting all the values we have
$t = \frac{1}{1.003 \cdot 10^{-6}-9.2568\cdot 10^{-6}} \ln \left( 0.27\cdot 0.4367 \frac{1.003}{9.2568}\right) \approx 5.2882\cdot 10^5$ seconds or $146.8951$ hours.

So the answer I get Is 14:00 22/4. The provided answer on the other hand claim I should get 17:00 25/4. Am I using the wrong method or the wrong chart or I did I make a mistake somewhere?

Last edited:
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#### haruspex

Homework Helper
Gold Member
2018 Award
Surely the activity ratio at time t (your R) is $\frac{\lambda_1N_1(t)}{\lambda_2N_2(t)}$.

#### Incand

Surely the activity ratio at time t (your R) is $\frac{\lambda_1N_1(t)}{\lambda_2N_2(t)}$.
Nice catch! Edited that now. I remembered the $\lambda$ in the next step in the equation however so it doesn't change the calculation itself.

#### haruspex

Homework Helper
Gold Member
2018 Award
Nice catch! Edited that now. I remembered the $\lambda$ in the next step in the equation however so it doesn't change the calculation itself.
I cannot find any flaw in your working. Interesting that the official answer is just about half.

#### haruspex

Homework Helper
Gold Member
2018 Award
different isobars
I guess you mean isotopes.

#### Incand

I cannot find any flaw in your working. Interesting that the official answer is just about half.
In that case I feel that at least I understand something:D I actually think I knew how they got the different answer now. Seems that if you calculate $t = \frac{1}{1.003 \cdot 10^{-6}-9.2568\cdot 10^{-6}} \ln \left( 0.27\cdot 0.4367\right)$m that is we actually take the quantities as the amount instead of the activity we get $71.94$ hours. So I'm guessing this may be an error in the answer since they say relative activity and mBq is a unit of activity. Or I guess it could just be a coincidence.

I guess you mean isotopes.
It should be, weirdly enough the question actually had isobars in it.

#### haruspex

Homework Helper
Gold Member
2018 Award
we actually take the quantities as the amount instead of the activity we get 71.9471.9471.94 hours
Yes, that must be it.

e question actually had isobars in it.
The perils of predictive text?

"Age of radioactive sample from sample"

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