Answer:981 .52 years
Explanation:
According to the first order kinetics for radioactive decay, the rate constant is given as
k = [tex]\frac{1}{t}[/tex] ㏑ [tex]\frac{initial activity}{final activity}[/tex]
making t subject formulae becomes
t= [tex]\frac{1}{k}[/tex] ㏑ [tex]\frac{initial activity}{final activity}[/tex]
but K = 0.693/ t1/2
using the half life, t 1/2 for carbon 14 = 5730 years since not given
therefore k = 0.693/ 5730 y
1.2 x 10 ^-4y-1
Bringing back our equation becomes
t= [tex]\frac{1}{k}[/tex] ㏑ [tex]\frac{initial activity}{final activity}[/tex]
where initial activity = 15.3 disintegrations per minute per gram for living organsism
and final activity= 13.6 disintegrations per minute per gram for leg bone
t = 1/1.2 x 10^-4y-1 ㏑( 15.3/ 13.6)
=981 .52years
The age of the leg bone is 981.52 years
The age of the bone sample of the animal is 976 years.
Using the relation;
0.693/t1/2 = 2.303/t log No/N
t1/2 = half life of carbon-14 = 5730 years
t = age of the sample = ?
No = Decay rate of carbon-14 in a living organism = 15.3 disintegrations per minute per gram
N = Decay rate of carbon-14 in the sample = 13.6 disintegrations per minute per gram
Substituting values;
0.693/5730 = 2.303/t log (15.3/13.6)
t = 976 years
The age of the bone sample of the animal is 976 years.
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