A by-product of some fission reactors is the isotope^239 _94 Pu, which is an alp

Question

A by-product of some fission reactors is the isotope^239 _94 Pu, which is an alpha emitter with a half-life of 24, 000 years.^239 _94 Pu rightarrow^235 _92 U +^4 _2 He Consider a sample of 1.2 kg of pure^239 _94 Pu at t = 0. (a) Calculate the number of^239 _94 Pu nuclei present at t = 0. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error, nuclei (b) Calculate the initial activity of the sample. Bq (c) How long does the sample have to be stored if a "safe" activity level is 0.12 Bq? Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. yr

Explanation / Answer

(a)Number of atoms in 1.20 kg sample of94239Pu at t = 0 is given by

[(1.2)/(234)] * (6.022 * 1026) = 3.0882 *1024

b)

Activity is just the decay rate and

   Ro = No (ln2 / T )     where Ro is theinitial rate and T is the halflife

      = 3.0882 *1024 ( ln2 / 24000 yr (3.1557x107 sec / yr) )

      = 2.826 x1012   decay / sec    or    2.826x 1012 Bq   ( a Bqis a decay per sec)

c)

The only thing you had to do there was convert the half life into seconds by using the seconds per year conversion.

Now,

    R = Ro (1/2)t/T        solve thisfor t by dividing by Ro and taking ln and you get

    t = T ln ( R /Ro ) / ln(1/2) = 24000 yr * ln(0.12 / 2.826 x 1012 ) / ln(1/2)

      = 1,066,000years   ( a little more than a million years)

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