An important physical quantity at radioactive decay is the half-life.
The half-life T1/2 of a radioactive decay is the time in which one-half of the radioactive nuclei has disintegrated.
The number radioactive nuclei remaining after time t :
- N(t) = N(0). (1/2)t/T1/2
N(0) the number radioactive nuclei at t = 0 s
N(t) the number radioactive nuclei at t = t s
t tijd
T1/2 half-life
- N(t) = N(o) e– λt
λ = ln 2/T1/2
λ decay constant
Remark : t en T1/2 have the same unit of time
Example
The half-life of I-131 is 8.0 days.
i Determine the percentage of the radioactive nuclei that is disintegrated after 6 days.
ii After what time is 80 % disintegrated ?
Solution I
i N(t) = N(0) (1/2)6/8 = N(0) 0.59
After 6 days, 59 % of the number of radioactive nuclei are stil there.
Therefore 41 % is disintegrated
ii N(t) = N(0) (1/2)t/T1/2
N(t)/N(0) = (1/2) t/T1/2
80 % disintegrate… 20 % remain
0.20 = (1/2)t/T1/2
0.20 = (1/2)t/T1/2
log 0.20 = t/T1/2 log (1/2)
-0.699 = (t/T1/2)( -0.301)
t/T1/2 = (-0,699/-0.301) = 2,32
t = 2.32 T1/2 = 2.32 8 = 18,56 days
After 19 days 80 % is disintegrated
Solution II
i N(t) = N(0) e – λt
λ = ln 2/T1/2
λ = 0.693/8 = 0.0866 ( T1/2 in days)
N(t) = N(0) e – (0.0866) 6
N(t) =N(0) e – 0,5196
N(t) = N(0) 0,59
After 6 days, 59 % of the number of radioactive nuclei are still there.
Therefore 41 % is disintegrated.
ii N(t) = N(o) e– λt
N(t)/N(0) = e–λt
0.20 = e–λt
ln (0.20) = – λt
-1.609 = – 0.0866 t
t = 18.58 days
After 19 days, 80 % is disintegrated