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Specific acticvity (activiy per gram) of a sample of .^(239)Pu and .^(240)Pu was found to be 6xx10^(9) dps. Given that t_(1//2) (Pu - 239) and t_(1//2) (Pu - 240) are 2.44xx10^(4) year and 6.58xx10^(3) year respectively, then calcualte the isotopic compostion of mixture. |
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Answer» Solution :Specific activity expresses that `1g` sample shows the activity `6xx10^(9)` dps. Let the mixture CONTAIN `a g Pu-239` and `b g Pu-240`, then `a+b = 1` ....(1) Now, separately for `Pu-239` and `Pu-240`, the RATE of DECAY is `r_(1)` and `r_(2)`respectively. Then, `r_(1) = lambda.N = (0.693)/(t_(1//2)) xx (a xx N_(A))/(239) dpsg^(-1)` `= (0.693xx a xx 6.023xx10^(23))/(2.44xx10^(4)xx 365xx24xx60xx60xx239)` `= 2.27xx10^(9)xx a dpsg^(-1)` SIMILARLY, `r_(2) = (0.693)/(t_(1//2)) xx (b)/(240) xx N_(A) dpsg^(-1)` `= 8.38xx10^(9)xx b dpsg^(-1)` But, `r = r_(1) + r_(2)` `2.27xx10^(9) a + 8.38 xx 10^(9)b = 6xx10^(9)` or `2.27a + 8.38b = 6` ...(2) Solving equations (1) and (2), `a = 0.39` or `39%` `b = 0.61` or `61%` |
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