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

Starting at

A = Ao e-λt  we rearrange to A/Ao  = e-λt 

Starting at

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: