Question

The rate of decay of particular isotope of Radium (in mg per century) is proportional to its mass (in mg). A 50mg sample takes one century to decay to 48mg. Ho0w long will it take before there are 45 mg of the sample?

Answer 1

Amount of Radium after `t` centuries is `50(24/25)^t`

It will take 2.6 centuries for the Radium to weigh 45mg.

Explanation:

Let us define the following variables:

`{ (x,"mass of Radium (mg)"), (t, "time (centuries)") :}`

Then

`-dx/dt prop x => dx/dt = -kx`

where `k` is the constant of proportionality. This is a First Order separable Differential Equation and we can separate the variables to get:

`int \ 1/x \ dx = int \ -k \ dt`

Which we can integrate to get:

`\ \ \ ln |x| = -kt + C`
`:. ln x = -kt + C` , as `x` is positive

We initially started off with `x=50` (mg) `=>x=50` when `t=0`, so we can substitute into the DE solution to get:

`ln 50 = C`

We are also told that `x=48` (mg) when `t=1` (century) so we can substitute into the DE solution to get:

`ln 48 =-k + ln50 => k = ln50-ln48 = ln(50/48)`
`:. k = ln(25/24)`

And so the Specific Solution is:

`ln x = -tln(25/24) + ln 50` ....[1]
`:. ln x = ln(25/24)^(-t) + ln 50`
`:. ln x = ln (50(25/24)^(-t))`
`:. x = 50(25/24)^(-t)`
`:. x = 50(24/25)^t`

[ We should just check that we have not made a mistake by checking the initial condition:

`t=0 => x=50(24/25)^0=50`
`t=1 => x=50(24/25)^1=24`

so we know the solution is sound]

We are asked to find `t` when `x=45`, and so using [1] we have:

`ln 45 = -tln(25/24) + ln 50`
`:. tln(25/24) = ln 50 - ln45`
`:. tln(25/24) = ln 50/45`
`:. tln(25/24) = ln 10/9`
`:. t = (ln 10/9)/(ln(25/24))`
`:. t = 2.58097 ...`

Hence it will take 2.6 centuries.