Question

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?

Answer 1

`200"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_(1/2)` that have elapsed:

`n_(1/2)=(2.82xx10^9)/(7.04xx10^8)=4`

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

Initial mass = `12.5 xx2xx2xx2xx2=200"g"`

1. General Method.

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

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

`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=0.693/(t_(1/2))`

`:.lambda=(0.693)/(7.04xx10^(8))=9.84xx10^(-10)" "a^(-1)`

Taking natural logs of both side of `color(red)((1))rArr`

`lnN_t=lnN_0-lambdat`

We can use mass in grams instead of atoms so:

`lnM_t=lnM_0-lambdat`

`:.lnM_0=lnM_t+lambdat`

`:.lnM_0=ln(12.5)+(9.84xx10^(-10)xx2.82xx10^9)`

`lnM_0=2.526+2.775=5.301`

`:.M_0=200.5"g"`