# 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?

Aug 15, 2016

$\textsf{8 \textcolor{w h i t e}{x} \text{hr}}$

#### Explanation:

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:

$\textsf{- {N}_{t} / t \propto {N}_{t}}$

Putting in the constant:

$\textsf{- {N}_{t} / t = \lambda t}$

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

$\textsf{{N}_{t} = {N}_{0} {e}^{- \lambda t} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{{N}_{t}}$ is the number of undecayed atoms at time $\textsf{t}$

$\textsf{{N}_{0}}$ is the initial number of undecayed atoms

$\textsf{\lambda}$ is the decay constant

$\textsf{t}$ is the time.

The data given will enable us to calculate $\textsf{\lambda}$ and, from that, the half - life.

Taking natural logs of $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{\ln {N}_{t} = \ln {N}_{0} - \lambda t}$

$\therefore$$\textsf{\ln \left({N}_{t} / {N}_{0}\right) = - \lambda t}$

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

$\textsf{\ln \left(\frac{1250}{10000}\right) = - \lambda \times 24}$

$\therefore$$\textsf{\ln \left(\frac{1}{8}\right) = - \lambda \times 24}$

$\textsf{- \lambda = - \frac{2.079}{24}}$

$\textsf{\lambda = 0.0866 \textcolor{w h i t e}{x} {\text{hr}}^{- 1}}$

We can find the 1/2 life by setting the condition that when $\textsf{{N}_{t} = {N}_{0} / 2}$ then $\textsf{t = {t}_{\frac{1}{2}}}$

Substituting these into $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{\frac{\cancel{{N}_{0}}}{2} = \cancel{{N}_{0}} {e}^{- \lambda {t}_{\frac{1}{2}}}}$

$\therefore$$\textsf{\ln 2 = \lambda {t}_{\frac{1}{2}}}$

$\textsf{{t}_{\frac{1}{2}} = \frac{0.693}{\lambda} = \frac{0.693}{0.0866} = 8 \textcolor{w h i t e}{x} \text{hr}}$

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:

$\textsf{10 , 000 \rightarrow 5000}$ is 1 half life

$\textsf{5000 \rightarrow 2500}$ is another half life

$\textsf{2500 \rightarrow 1250}$ is another half - life.

So a total of 3 half - lives has elapsed.

$\therefore$ 3 half - lives = 24 hrs

$\therefore$ 1 half - life =24/3 = 8 hrs.