Question

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 hours (3) 3 × 12.4 hours(2) 2 × 12.4 hours (4) 4 × 12.4 hours

Answer 1

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

Answer 2

Answer:

The answer is (3) 3 × 12.4 hours