# Nuclear physics: determine the date of the Chernobyl accident

## Homework Statement

The acitvities from the fission products 131I and 133I were measured in the air of Gothenburg April 28 1986 at 17:00. The result was 0.12 Bq/m3 and 0.39 Bq/m3 for 131I and 133I respectively. These isotopes came from the Chernobyl nuclear reactor disaster. Use this information to calculate when the reactor container exploded. The relative amount of 131I and 133 produced in the fission of 236U is 2.892 and 6.686 percent respectively.

## Homework Equations

t½(131I) = 8 days = 24 x 60 x 60 x 8 seconds = 691200 sec
t½(133I) = 21 hours = 21 x 60 x 60 seconds = 75600 sec

The activity for short-lived nuclides: $$A(t)=\lambda N_{0}e^{-\lambda t}$$, where $$A$$ is the acitivity, $$N$$ the number of radioactive nuclei and $$N_{0}=N(t=0)$$

## The Attempt at a Solution

The decay law, numerical values inserted for $$^{131}I$$ and $$^{133}I$$ respectively, divided by eachother to get rid of $$N_{0}$$ which is unknown:

$$\frac{2.892 \times 0.12}{6.686 \times 0.39}=\frac{e^{-\lambda_{131}t}}{e^{-\lambda_{133}t}}$$

Some algebra gives $$t=137707$$ seconds. Subtracting this from the given date gives April 27 02:44:53 as the date of the Chernobyl disaster. Wikipedia (for instance) states that the accident happened "26 April 1986 01:23:45 a.m. (UTC+3)" which is the same as April 26, 03:23:45 a.m Gothenburg time. Since my solution is so far off from the actual date I figured I must have done something wrong. Can anybody help me out here, please?

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malawi_glenn
Homework Helper
Dear uppsala student

the equations you can have is:

0.12 = N_o(131) Lambda(131) exp(- Lambda(131) T )

0.39 = N_0(133) Lambda(133) exp(- Lambda(133) T )

and at T = 0:
(N_o(131))/(N_o(133)) = 2.891/6.686

Right?

Now try again

Thanks for you help! I got it right this time!

malawi_glenn
Homework Helper
Thanks for you help! I got it right this time!
Great, good luck on the exam