Question

The half life of iron-59 is 44.5 days. After 133.5 days, 2.76 g of iron-59 remains. What was the mass of the original sample?

How do I find the mass? I looked through the book and I didn't see it

Answer 1

The mass of the original sample was `22.08` `g`.

Explanation:

We can describe exactly what's happening here using a fractional exponential function. Start by noticing that after `color(blue)44.5` days, the value is halving

Let `color(red)t` be the elapsed time in days.
Let `N` be our initial mass.
Let `M` be our final mass.

`M = N*(1/2)^(color(red)t/color(blue)44.5)`

A simple way to interpret this is to first ignore the exponent to recognize that the mass is halving. Next, looking at our exponent, we see that when `color(red)t=color(blue)44.5`:

`(1/2)^(color(red)t/color(blue)44.5) = 1/2`

This is exactly what we would expect! After the length of the half-life has passed, the mass has halved.

You will notice that we can easily swap `color(blue)44.5` with some other half-life and our calculation will still be valid.

To solve this question, we are given the final mass and must find the initial mass. Unfortunately, our formula gives us the final mass in terms of the initial mass, so we need to rearrange it.

`M/((1/2)^(t/44.5)) = N`

`M*(1/2)^(-t/44.5) = N`

`color(green)M*(2)^(color(red)t/44.5) = N`

`color(green)2.76*(2)^(color(red)133.5/44.5) = N`

`2.76*2^3 = N`

`2.76*8 = N`

`22.08 = N`

Therefore, the mass of the original sample was `22.08` `g`.

Answer 2

Radioactive decay is a first-order process. Consider,

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

Let's derive the rate constant,

`44.5d = 0.693/k`
`therefore k approx 1.56*10^-2d^-1`

Moreover, consider the equation for first-order processes,

`ln[A]_t = -kt + ln[A]_0`
`=> ln(([A]_t)/([A]_0)) = -kt`

Hence,

`ln(([A]_t)/([A]_0)) = -kt`

`therefore [A]_0 approx 22.1g`