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
                          Â
k= 0.6931/29.1
                          Â
k= 0.02382 (1/ year)
                          Â
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
