### 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* Δ*t*

where* P *is the probability and* *Δ*t *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 = N**_{o }**e**^{-λt}

Where N is the number of original atoms remaining after time t and N_{o} was the number at the start

Since the amount of activity is directly proportional to the number of atoms available for decay, then:

**A = A**_{o }**e**^{-λt}

Where A is the activity after time t and A_{o} was the activity at the start

Starting at

**A = A**_{o }**e**** ^{-λt }**we rearrange to

**A/A**

_{o}**= e**

^{-λt }Starting at

**A = A**_{o }**e**** ^{-λt }**we rearrange to

**A/A**

_{o}**= e**

^{-λt1/2}where **t _{1/2}** is the half life.

Taking logs on both sides we get

** ln2 = λt**

The value of ln2 is 0.693 so finally:

** 0.693/t _{1/2} = 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 = A**_{o }**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/t _{1/2} = λ**

so **λ = o.693/1 = 0.693** per year.

from **A = A**_{o }**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)**

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