In 2006, a former Russian spy was killed via poisoning with 10.0g of 210 Poloniu

Question

In 2006, a former Russian spy was killed via poisoning with 10.0g of210Polonium, which decays via emission and has a half life of 138 days.

Explain why poisoning with an alpha emitter is a “perfect crime” (very lethal and very hard to detect)

The spy died 10 days after the initial poisoning. What percentage of the original 210Po remained in his system when he died?

Calculate the amount of energy that was released from the nuclear disintegration of 210Po to its product. You will need to write a balanced equation and look up the precise mass of the reactants and products at https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html (click on a nuclide to see the data in the table on the bottom of the page; scroll to the right).

Explanation / Answer

Radioactive decay follows first order kinetics and using the integrated rate expression for first order

    K(lambda) = 2.303 log (a/a-x) /t   where k is the rate constant , time is t and a is initial concnetration and a-x is the amount remaining after time t

Given half life = 138 days thus K = 0.693/136 per day

time = 10 days and a = 10 micrograms = 10-5 g

substituting and calculating

a-x =9.5 x10-6 g = 9.5 micrograms

percentage of polonimum remaining after 10 days = 95%

disintegration of polonium is 84Po210 -----> 82Pb206 + 2He4

mass defect = [205.9744653 + 4.00150] - 209.982848 =-0.0068827 u

energy corresponding to this mass is = 0.0068827 x 931.494028 MeV =6.411193946 Mev

given

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.