### Half life and the decay constant

These notes support the video shown at the bottom of this page.

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: