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 99.9352% of the 38Sr90 released in a nuclear reactor accident to disappear?

Answer 1

Initial amount of 38Sr90 = N₀

99.9352% of 38Sr90 has disappeared therefore only 0.0648% of the initial amount is remaining.

Amount of 38Sr90 remaining = N = (6.48x10^-4)N₀

Time taken for this amount of 38Sr90 to remain = T

Half-life of 38Sr90 = T_1/2 = 29.1 years

Decay constant =

= 2.382 x 10^-2 years^-1

Taking natural log on both sides,

T = 308.2 years

Time taken for 99.9352% of the 38Sr90 released to disappear = 308.2 years