The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
![(1/2) ^{n} =x](https://tex.z-dn.net/?f=%281%2F2%29%20%5E%7Bn%7D%20%3Dx)
![t_{1/2} = \frac{t}{n}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7Bt%7D%7Bn%7D%20)
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
![t_{1/2}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20)
- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
![(1/2) ^{n} =x](https://tex.z-dn.net/?f=%281%2F2%29%20%5E%7Bn%7D%20%3Dx)
</span>
![(0.5) ^{n} =0.125](https://tex.z-dn.net/?f=%280.5%29%20%5E%7Bn%7D%20%3D0.125)
![n*log(0.5)=log(0.125)](https://tex.z-dn.net/?f=n%2Alog%280.5%29%3Dlog%280.125%29)
![n= \frac{log(0.5)}{log(0.125)}](https://tex.z-dn.net/?f=n%3D%20%5Cfrac%7Blog%280.5%29%7D%7Blog%280.125%29%7D%20)
![n=3](https://tex.z-dn.net/?f=n%3D3)
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
![t_{1/2} = 12.4](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%2012.4%20)
</span><span>
![t_{1/2} *n = t](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%2An%20%3D%20t%20)
</span>
![t= 12.4*3](https://tex.z-dn.net/?f=t%3D%2012.4%2A3%20)
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>