90Sr has a half life of 28.5y. It is chemically similar to Ca and enters the body through the food chain and collects in bones. It is a serious health hazard. How long in (years) will it take for 99.99% of Sr released in a nuclear reactor accident to disappear?
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If initial mass is $M$. Amount remaining after $n$ half lives is $M/2^{n}$. Ask yourself, How much remains after 99.99% is gone? Equate. |
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Pratik's point was a good one, though perhaps not the most helpful. The decay of a radioactive material means that the amount of time for any specific percentage of a quantity of material to decay is constant. That is, in this case, $0.0001 N_0 = N_0 e^{-\lambda t}$ $t = -\frac{\ln(0.0001)}{\lambda}$ where $\lambda$ is called the decay constant. In this case we have $t \approx 379 \mbox{years}$ What you really should care about is the effect of this material. When making these sorts of calculations for accident analysis, engineers consider, a huge variety of factors such as:
There is an entire field known as probabilistic risk assesment that examines these sorts of questions and debates the relative weights that should be placed on them. I am not a PRA expert, but I have some experience with the field. That being said, we, unfortunately, have some real data with which we can work. In the case of Fukushima, the level of ${}^90$Sr in seawater some 7 months after the accident was less than $0.03 Bq/cm^3$; see this article. Another source gives a value of 195 Bq. This equates to about $4 \times 10^{-11}$ grams... not very much. Considering that the deposition fraction (the amount that gets into bones once consumed/inhaled) is about 0.12, these quantities are quite minimal. In short, it's not something you need to lose sleep over. |
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