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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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2 Answers 2

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:

  • amount of fuel in reactor,
  • original composition of fuel,
  • history of fuel (reactor operation),
  • relative liklihood of different types of accident,
  • relative liklihood of the release of refractory products (material from the core),
  • likely percent of refractory products released to the environment,
  • relative liklihood of dispersion of various refractory products (considering half-life, air bouyancy, diffusion factors, etc.),
  • potential containment measures and all of the possibilities that go into their success or failure,
  • decay chains of environmental contaminants
  • environmental vectors for human exposure
  • relative effect of various levels of exposure
  • relative effect of type (inhalation, consumption, skin, etc.) of exposure
  • etc., etc.

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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If the half-life is 28.5 yr, then your 1.5 d can't be right. I think you've calculated the time for 0.01% of the sample to decay. Pratik's answer would give 13.3 half-lives or 379 years. – Warrick Oct 15 '11 at 19:02
Ah yes, sorry. Fixed now – AdamRedwine Oct 16 '11 at 11:26

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?


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I think the poster was hoping to get an answer that accounted for the determination of some real value of M. – AdamRedwine Oct 14 '11 at 19:19

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