Half life and the decay constant
The radioactive decay constant is represented by the symbol λ. The definition of λ is the probability that any particular nucleus will decay in a unit of time. As an equation, that is:
P = λ Δt
where P is the probability and Δt is a time much less than the half life.
The radioactive decay Law states that this probability is constant
Derived from the Law is the expression:
N = No e-λt
Where N is the number of original atoms remaining after time t and No was the number at the start
Since the amount of activity is directly proportional to the number of atoms available for decay, then:
A = Ao e-λt
Where A is the activity after time t and Ao was the activity at the start
A = Ao e-λt we rearrange to A/Ao = e-λt
A = Ao e-λt we rearrange to A/Ao = e-λt1/2
where t1/2 is the half life.
Taking logs on both sides we get
ln2 = λt
The value of ln2 is 0.693 so finally:
0.693/t1/2 = λ
Calculating future radioactivity using the decay constant
Knowing a current value of activity we can easily calculate itsʼ value several half lives later by halving that value several times: For example if the activity now is 800 counts per second and the half-life is one year then:
- Now 800
- 1yr 400
- 2yr 200
- 3yr 100
- 4yr 50
and so on, similarly, if the specimen we have is several years old we can work out that the activity:
- one year ago was 1600
- two years ago 3200,
- three years ago 6400 and so on.
The maths is a bit more difficult if we want to calculate the activity over a fractional number of half lives, for example in 6 months or 2.3 years ago, but we can do it. Let’s use the same example. If the activity is 800 counts per second and the half life 1 year, then after si months we know the activity must have fallen but not so far as 400 because that is the likely activity after 1 year. The answer is not 600 because it is not a linear relationship but it might not be too far from that.
To do this we use the equation: A = Ao e-λt
where A is the activity at time t, Ao is the activity at the start, λ is the decay constant. Remember that:
0.693/t1/2 = λ
so λ = o.693/1 = 0.693 per year.
from A = Ao e-λt so A = 800 x (2.718 to the power of -0.693 x 0.5) (0.5 because 6 months is half of a year)
because of the – sign that is 800 x 1/(2.718 to the power of 0.693 x 0.5)
= 800 x 0.717 = 574 counts per second (to 3 significant figures)
Other pages which are closely related to this are: