Question

How do you express the concept of the rate of decay of a radioactive substance as a differential equation?

When a radioactive substance decays, the rate of decay is proportional to the amount of the substance remaining.
a) Express this fact as a differential equation.
b) Solve the differential equation to find an equation for m(t), the amount of mass remaining after
time t.
c) Suppose the time is measured in days. There are initially 125.3 grams of the substance. After 3
days, 98.1 grams remain. Use these facts to find the constant k for this substance.
d) Using the value for k that you found, predict the amount of the substance that will remain after 6
days of decay

Answer 1

See below.

Explanation:

From the question the rate of the decay of the mass is proportional to the mass remaining at time `t`. And because the mass is decreasing with time the constant of proportionality will be negative and thus the differential equation can be expressed as ,

.`[a]`........`[dm]/dt=-km`, where `m`=mass and `k` is the constant of proportionality. we now need to integrate this expression,
inverting both ides of .......`[a]`,

`[dt]/[dm`=`-1/[km]`, integrating w.r.t `m`, `t=-1/klnm+C`......`[2]`.

Now let `m=m0` [ the initial mass when `t=0`]

So from .....`[2]` `0=-1/klnm0+C`, therefore `C=1/klnm0`

`t=-1/klnm+1/klnm0`, `t=1/k[lnm0-lnm]`

Therefore, `kt=ln[[mo]/[m]]` and so, `e^[kt]=[mo]/m` which will give `m=m0e^[-kt]`....and this is the answer to part `b`.

For part ...`[c]` we are given that `m0` [ intial mass =`125.3] and`m=98.1[ mass remaining when `t=3`]

And so from answer part `[b]`, `98.1=[125.3]e^[-3k]`

`98.1/125.3=e^[-3k]`, therefore -`1/3ln[98.1/125.3]`=k [ theory of logs]

When evaluated the above expression yields a value of `k`=`-0.081575`[approx answer to part `c`]

For part `[d]`, we now know that `m` =`m0e^-[0.081575t]` and we are given that `t=6` and we know the initial amount is `125.3` grammes.

Therefore `m=125.3e^-[6[0.081575]]` and when evaluated gives `m` the amount remaining =`76.8043` grammes[ approx]

Hope this was helpful.