Question

Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and collects in bones. Consequently, 3890Sr is a particularly serious health hazard. How long (in years) will it take for 99.99% of the 3890Sr released in a nuclear reactor accident to disappear?

Answer 1

It will take approximately 378.70 years for 99.99% of the strontium-90 released in a nuclear reactor accident to disappear, calculated based on its half-life of 28.5 years.

Step-by-step explanation:

To determine how long it will take for 99.99% of the strontium-90 (90Sr) to disappear, we can use the formula for exponential decay based on the half-life of the isotope. The half-life of 90Sr is 28.5 years.

The formula to calculate the number of half-lives (n) needed for a certain percentage of a radioactive substance to remain is given by:

n = log(final amount/initial amount) / log(0.5)

For 99.99% to disappear, we want the final amount to be 0.01% of the initial amount. Substituting these values into the formula provides:

n = log(0.0001) / log(0.5) = 13.2877 half-lives

Then, to find the total time in years, we multiply the number of half-lives by the half-life of the isotope:

Total time = n × half-life of 90Sr

Total time = 13.2877 × 28.5 years = 378.70 years

So, it will take approximately 378.70 years for 99.99% of the 90Sr released in a nuclear reactor accident to disappear.