Preparation problems

43-36) If radon infiltration into a house were stopped, what fraction of the initial radon would be left after (a) 1 day (b) 1 week or (c) 1 month?

N = N_o 2^-t/t_1/2
t_1/2 = 3.82 days (from table 43-2)
(a) N(1 day) = N_o 2^-1/3.82 = 0.834 N_o so 83.4% of original remains after 1 day
(b) N(1 week) = N_o 2^-7/3.82 = 0.281 N_o so 28.1% of original remains after 1 week
(c) N(1 month) = N_o 2^-30/3.82 = 0.004 N_o so 0.4% of original remains after 1 month

43-42) Below are reported levels of iodine-131 contamination in milk for several countries affected by the 1986 Chernobyl accident, along with each country's safety guideline. Given that I-131's half life is 8.04 days, how ong did each country have to wait for I-131 levels to decline to the level considered safe?

R(t) = R_o 2 ^-t/t_1/2
ln( R(t)/R_o ) = - (t/t_1/2) ln2
t = - t_1/2 [ ln ( R(t)/R_o ) ] / ln 2
= ( - 8.04 days / 0.693 ) ln ( R(t)/R_o )
= ( - 11.6 days ) ln ( R(t)/R_o )
Poland: t = ( - 11.6 d ) ln (1000/2000) = 8.04 days

Sweden: t = ( - 11.6 d ) ln (2000/2900) = 4.31 days

Austria: t = ( - 11.6 d ) ln (370/1500) = 16.23 days

Germany: t = ( - 11.6 d ) ln (500/1184) = 10.0 days

43-65) Some human lung cancers in smokers may be caused by polonium-210, which arises in the decay series of uranium-238 that occurs naturally in fertilozers used on tobacco plants. Wrtie the equations for (a) the procuction and (b) the decay of Po-210 (c) How might the health effects of Po-210 differ if its half-life were 1 day or 10 years instead of the actual 138 days?

(a) According to Fig 43-19, Po-210 is a decay product of Bi-210 (from b decay), so the equation is
²¹⁰₈₃Bi -> ²¹⁰₈₄Po + e^- + v
(b) Also from Fig 43-19, Po-210 decays by a decay into Pb-206, so the equation is
²¹⁰₈₄Po -> ²⁰⁶₈₂Pb + ⁴₂He
(c) If you look at the effect of one cigarette, the helath effects would be much better for a different half-life. A half life of 1 day would mean that much of the Po would decay before the cigarette was even smoked. A half-life of 10 years would mean that there would bevery little radiation, since R = lN and a big half-life makes a small l. On the other hand, for a regular smoker, it will probably not make much difference either way. Because the smoker is constantly replenishing the Po in his/her lungs, the Po level would reach some equilibrium value. If the half-life were very short, there would not be much Po in the lung at any given time, but with the large decay constant the activity would still be fairly high. If the half-life were very long, then there would be a large amount of Po built up, but the activity would be about the same since the decay constant is smaller for this case. The bottom line is that it is hard to change R. If l is big, then N is small, but if l is small, then N gets big, so the product of the two is more or less constant and depends more on how many cigarettes he smokes rather than the half-life.

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Solutions translated from T_EX by T_TH, version 1.57.