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 99.9488% of the 38Sr90 released in a nuclear reactor accident to disappear?

Answer 1

Given:

Strontium 38Sr90

Half life= 29.1 year

99.9488% of the 38Sr90 released in a nuclear reactor accident to disappear?

Decay constant , k = ln(2)/half life

                            k= 0.6931/29.1

                            k= 0.02382 (1/ year)

                            k=0.02382 (year^-1)

We know,

ln(A/Ao) = -kt

A = amount remaining at time t

Ao=initial amount

99.9488% disappears

100 - 99.9488 = 0.0512 % remains

(A/Ao) x 100% = 0.0512%

(A/Ao) = 0.00512

ln(0.00512) = -0.02382 x t

Time t = 221.4 years