Download Half Life and the Decay Constant

Key Stage 5 PHYSICs I Lesson Plan 5 – Nuclear medicine Resource Sheet 5.2
Half Life and the Decay Constant
If we have a large number of nuclei of a single species statistics can be used to
make predictions about the average number of decays occurring in a small time
interval.
Rolling a dice can be thought of as an analogy for nuclear decay. For example, if
rolling a 3 represents nuclear decay, and results in the die’s removal, but all other
outcomes do not, then we would expect, on average the change in the number
of dice removed, DN, to be 1/6 of the total number of dice (N) present. Or as an
equation:
DN = - –1 N
6
(negative since dice are removed).
But the 1/6 only comes in because of the rules of the game, if we instead chose
to remove 2s and 3s the constant would become 1/3.
•
How could you change the rules to make it ½? To make it 5/6?
Also the rule above is for one throw only. If we have several throws, then
we would apply the rule at each throw. For real radioactivity time passes
continuously, so we say:
DN
= – lN
Dt
eqn. 1
Where l is what is known as the decay constant. It represents the probability of a
nucleus decaying in the time interval Dt and takes the part of the fraction 1/6 in
the dice analogy.
Equation 1 can be written in calculus notation as
dN
= – lN
dt
eqn. 1a
Which can be integrated to give
N = N0e-lt
Where N0 is the number of nuclei present at time t=0.
eqn. 2
Key Stage 5 PHYSICs I Lesson Plan 5 – Nuclear medicine Resource Sheet 5.2
Rearranging eqn. 2 gives
ln
N
N0
= – lt
Which can again be rearranged to give
ln N = ln N0 – lt
eqn. 3
Half Life
We often define a particular radioactive isotope by the time taken for half of
the nuclei in a given sample to decay. This property is called the half life and is
written T1/2.
This implies that
N
= 2–1 = e-lT
N0
1
or ln 2– = – lT
1/2
1/2
And this leads to the equation
T = 0.693
l
eqn. 4
1/2
The average lifetime, Ta, of any nucleus may be calculated by multiplying lN(t)dt,
the number of nuclei disintegrating in the time interval dt, by the time t for
which the decaying nuclei exist, summing these products over all nuclei and then
dividing by the total number of nuclei at the start.
Therefore the average lifetime of a single nuclei can be expressed as
T=
a
1
N0
∞
∫ tlN e
0
0
dt = 1
l
-lt
This is not surprising, since in the dice analogy given above l was 1/6. You would
expect a die, on average, to last 6 throws between each showing of a 3.
eqn. 5