Question

A 0.125 mg sample contains a mixture of 239Pu and 240Pu in unknown proportions. The activity of the sample is 7.25 x 10^5 Bq. Calculate the wt % of ech Pu isotope.

Answer 1

Specific activity of ²³⁹Pu = 0.063 Ci/g

1 Ci = 3.7 x 10¹⁰ dps

so 0.063 Ci/g corresponds to 0.063 x 3.7x10¹⁰ = 0.23 x 10¹⁰ Bq = 2.33x10⁹ Bq or 2.33x10⁶ Bq/ g

Specific activity of ²⁴⁰Pu =0.23 Ci/g

1 Ci = 3.7 x 10¹⁰ dps

so 0.23 Ci/g corresponds to 0.23 x 3.7x10¹⁰ = 0.851 x 10¹⁰ Bq = 8.51x10⁹ Bq/ g or 8.51x10⁶ Bq/ g

Given Data

Sample weight : 0.125 mg

Amount of activity = 7.25 x 10⁵ Bq

let x be the weight fraction of ²³⁹Pu and the rest is ²⁴⁰Pu

(0.125x)(2.33x10⁶) + 0.125(1-x)(8.51x10⁶ )= 7.25x10⁵

x0.291x10⁶ +1.063x10⁶ - 1.063x10⁶ x = 7.25x10⁵

-0.772 x10⁶ x = -3.38 x 10⁵

x = 3.38 x 10⁵ /0.772 x10⁶ = 0.437

So ²³⁹Pu is 43.7% and ²⁴⁰Pu is 56.3% in the given sample