Question

Question 3.210Po is a very toxic isotope (about 106 times more toxic than cyanide). The main hazard is its intense alpha radiation which, if ingested, does serious damage to internal tissues. Imagine that you are an undercover spy and you want to poison some critics of your government by slipping some 210 Po into their margaritas. The maximum safe body burden of 21°Po is ~1000 Bq = 16.7 dpm. To make sure your deadly poison works you want to increase the body burden by >10 times to >167 dpm. Assume secular equilibrium. a) Your chemical lab knows how to separate 210 Po from 210Pb. You just have to get them the material. You have 200 dpm of pure 210Pb. How long do you have to let the 210Pb decay so that you have almost (95%) 200 dpm of 210Po? The half-life of 210Pb is 22.3 days. (5 pts) b) Your lab chemists do their job well and now you have a sugar cube containing 200 dpm of 210Po. You get ready to meet your victim in a bar in Five Point but suddenly he is out of the country (gone to Britain) for 138 days. Finally you meet and while his back is turned you slip the sugar cube into his drink. Will your poison work? The half-life of 210Po is 138 days. (5 pts)

Answer 1

Well, I must say it is a very curious question (Curie must be shining more brightly in her grave).

This question is an application of first order kinetics. Polonium-210 is a very toxic isotope as it undergoes alpha decay giving the following reaction which follows the first order kinetics:

Page : Date: / / 206 Po 82 + 14 He 212 84 X-radiation which is responsible for the toxicity of po. Radioactive decay of solo follows first order Kinetics. 1 i.e. -dclo] = 1 concentration to Tat timet rate constant or decay constant à in this case. => -d[po] - [lol Tro at On Integration : I In [P = - t - EPolo In Clolo - Attc that At t=0 [lo] = [lol => In 1= x(0) + - E) ICEOL c At = Infpol. [ho]

Page : Date: / / 20= Similarly, formation tollows first order of Kinetics. => t = ln [Pb to tot -10 210 (a) We have alb such that we or the zlopb and we want lo have asy. 21 . left should be 5 % I now halt lite is given : Half life is the time in which half of the reactant gets converted to product. i.e. for Pb , at half life ty +[lb= ['b} => xty = ln ! typ = 0.693 - ② 2 We can know & from ② d = 0.693 - 22.3 days - = 0.031 days' - ③

Page Date: / 2TO to lo Pb - t=0 200 dom tot 5 x 200 100 Pb lett 95 x 200 100 at time t = 5x200 û Substituting 0 and 9 in to.Ogi dan in 10 ( 39.2001 - ning x lm (0.08 ) : 96.65 days TOU It pb is for 96.65 converted to kept to | -97 days) 95% of decays wit will to: get lo has to work as a poison I slot should be > 167 dpm. we have ty => & can be calculated = 0.693 - 138 days Let us Calculate the [po ] after 138 day from egr 0 xt = In [lot. - [lo]

Page : Data / 138 In [Polo - At = ln [po] => In (200) - 0.693 x 188 = Incpol In [Pol=4.605 [Po é 4605 - 99.98 difius => [po] < 167 down 1 => It will not act as a poison

The concentration of Po after 138 days was to be 99.98 dpm which is less than the concentration required i.e. 167 dpm to make Po a poison.

So, our mission is unsuccessful but we give life to someone.