Question

If a scientist purifies 10 gram of radium-226, how many years must pass before only 0.50 gram of the original radium-226 sample remains unchanged?

Answer 1

The number of years is `=6911"years"`

Explanation:

The half life of `"Radium"-226` is `t_(1//2)=1599"years"`

The radioactive constant is `lambda=ln2/t_(1//2)`

The equation for radioactive decay is

`(N(t))/(N_0)=e^(-lambdat)`

Therefore,

`e^(-lambdat)=(N(t))/N_0=0.5/10=0.05`.

Taking natural logs of both sides,

`-lambdat=ln(0.05)`

`t=- (ln 0.05)/(lambda)`

`=- ln0.05/(ln2//t_(1//2))`

`=- ln0.05/ln2*t_(1//2)`

`=- ln0.05/ln2*"1599 years"`

`=` `" 6911years"`