FE Physics
Formulas & Explanation

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 :

1.       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

2.       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