Question

the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​

Answer 1

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$