Question

strontium 38sr90 has a half-life of 29.1 yr. it is chemically similar to calcium, enters the body through the food chain, and collects in the bones. consequently, 38sr90 is a particularly serious health hazard. how long (in years) will it take for 99.9778% of the 38sr90 released in a nuclear reactor accident to disappear?

Answer 1

It will take approximately 135.4 years for 99.9778% of the 38Sr90 released in a nuclear reactor accident to disappear.

Step-by-step explanation:

The half-life of 38Sr90 is given as 29.1 years. To find out how long it will take for 99.9778% of the 38Sr90 to disappear, we need to calculate the number of half-lives required. Since the final amount of 38Sr90 is 0.0222% (100% - 99.9778%), we can use the formula:

Final amount = Initial amount * (1/2)(number of half-lives)

By substituting the values and solving the equation for the number of half-lives, we find that it will take approximately 4.67 half-lives for 99.9778% of the 38Sr90 to disappear. Multiplying this by the half-life of 29.1 years gives us the answer: it will take approximately 135.4 years for 99.9778% of the 38Sr90 released in a nuclear reactor accident to disappear.