# If the half life of uranium-235 is 7.04 * 10^8 y and 12.5 g of uranium-235 remain after 2.82 * 10^9 y, how much of the radioactive isotope was in the original sample?

Jan 20, 2016

$200 \text{g}$

#### Explanation:

1. Quick Method.

Examiners and text - books often give nice numbers where the answer drops out easily.

In this case you count the number of 1/2 lives ${n}_{\frac{1}{2}}$ that have elapsed:

${n}_{\frac{1}{2}} = \frac{2.82 \times {10}^{9}}{7.04 \times {10}^{8}} = 4$

Now multiply back from 12.5 by doubling it successively 4 times:

Initial mass = $12.5 \times 2 \times 2 \times 2 \times 2 = 200 \text{g}$

1. General Method.

If the numbers are not so nice you need to use the general equation for radioactive decay:

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

$\lambda$ is the decay constant

${N}_{0}$ = the initial number of undecayed atoms

${N}_{t}$ = the number of undecayed atoms at time $t$

We can get $\lambda$ from the 1/2 life using this expression:

$\lambda = \frac{0.693}{{t}_{\frac{1}{2}}}$

$\therefore \lambda = \frac{0.693}{7.04 \times {10}^{8}} = 9.84 \times {10}^{- 10} \text{ } {a}^{- 1}$

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

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

We can use mass in grams instead of atoms so:

$\ln {M}_{t} = \ln {M}_{0} - \lambda t$

$\therefore \ln {M}_{0} = \ln {M}_{t} + \lambda t$

$\therefore \ln {M}_{0} = \ln \left(12.5\right) + \left(9.84 \times {10}^{- 10} \times 2.82 \times {10}^{9}\right)$

$\ln {M}_{0} = 2.526 + 2.775 = 5.301$

$\therefore {M}_{0} = 200.5 \text{g}$