Nuclear physics: determine the date of the Chernobyl accident

[Bapelsin]

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

t_½(¹³¹I) = 8 days = 24 x 60 x 60 x 8 seconds = 691200 sec
t_½(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 each other 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?

Thanks in advance!

 

[malawi_glenn]

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

 

[Bapelsin]

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

 

[malawi_glenn]

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

Great, good luck on the exam