The answer is (3) 3 × 12.4 hours
To calculate this, we will use two equations:
\((1/2) ^{n} =x\)
\(t_{1/2} = \frac{t}{n}\)
where:
n - number of half-lives
x - remained amount of the sample, in decimals
\(t_{1/2}\) - half-life length
t - total time elapsed.
First, we have to calculate x and n. x is 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 × 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
\((1/2) ^{n} =x\)
\((0.5) ^{n} =0.125\)
\(n*log(0.5)=log(0.125)\)
\(n= \frac{log(0.5)}{log(0.125)}\)
\(n=3\)
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
\(t_{1/2} = 12.4\)
\(t_{1/2} *n = t\)
\(t= 12.4*3\)
Therefore, it must elapse 3 × 12.4 hours before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope
