If 1250 counts of an originally 10,000 count radioactive sample are being emitted after one (24 hour) day, what is the half-life of the element?

Formal Method:

When an atom decays this is a random event which obeys the laws of chance.

The greater the number of undecayed atoms in a sample, the more chance there is of one of them decaying. We can, therefore, say that the rate of decay is proportional to the number of undecayed atoms:

#sf(-N_t/tpropN_t)#

Putting in the constant:

#sf(-N_t/t=lambdat)#

By doing some integration, which I won't go into here, we get the expression for radioactive decay:

#sf(N_t=N_0e^(-lambdat)" "color(red)((1)))#

#sf(N_t)# is the number of undecayed atoms at time #sf(t)#

#sf(N_0)# is the initial number of undecayed atoms

#sf(lambda)# is the decay constant

#sf(t)# is the time.

The data given will enable us to calculate #sf(lambda)# and, from that, the half - life.

Taking natural logs of #sf(color(red)((1))rArr)#

#sf(lnN_t=lnN_0-lambdat)#

#:.##sf(ln(N_t/N_0)=-lambdat)#

The count rate is proportional to the number of undecayed atoms so this becomes:

#sf(ln(1250/10000)=-lambdaxx24)#

#:.##sf(ln(1/8)=-lambdaxx24)#

#sf(-lambda=-2.079/24)#

#sf(lambda=0.0866color(white)(x)"hr"^(-1))#

We can find the 1/2 life by setting the condition that when #sf(N_t=N_0/2)# then #sf(t=t_(1/2))#

Substituting these into #sf(color(red)((1))rArr)#

#sf(cancel(N_0)/2=cancel(N_0)e^(-lambdat_(1/2)))#

#:.##sf(ln2=lambdat_(1/2))#

#sf(t_(1/2)=0.693/lambda=0.693/0.0866=8color(white)(x)"hr")#

This is an example of 1st order exponential decay and is common in nature.

Intuitive Method:

In questions the numbers often drop out nicely and they can be solved intuitively.

The time taken for the count rate to fall by half its initial value is equal to 1 half - life.

So:

#sf(10,000rarr5000)# is 1 half life

#sf(5000rarr2500)# is another half life

#sf(2500rarr1250)# is another half - life.

So a total of 3 half - lives has elapsed.

#:.# 3 half - lives = 24 hrs

#:.# 1 half - life =24/3 = 8 hrs.