Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

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Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

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Answer 1 · 2017-05-19T17:20:53

$7$ $7$ half-lives

$0.24 g$ $0.24g$ $^{210} P o$ $"^(210)Po$

Explanation:

The number of half-lives is simply the number of days given ( $966 d$ $966d$ ) divided by the half-life ( $138 d$ $138d$ ), which is $7$ $7$ half-lives.

The amount of $^{210} P o$ $"^(210)Po$ remaining after $t$ $t$ days is represented by the equation

$N (t) = N_{0} {(\frac{1}{2})}^{\frac{t}{t_{\frac{1}{2}}}}$ $N(t) = N_0(1/2)^(t/(t_(1/2))$

where $N (t)$ $N(t)$ is the amount of substance remaining after $t$ $t$ days,
$N_{0}$ $N_0$ is the initial amount of substance,
$t$ $t$ is the time in days, and
$t_{\frac{1}{2}}$ $t_(1/2)$ is the half-life of the substance.

Plugging in known variables, we have

$N (t) = (31 g) {(\frac{1}{2})}^{\frac{966}{138}} = 0.24 g$ $N(t) = (31g)(1/2)^((966)/(138)) = 0.24g$ $_{210} P o$ $"_(210)Po$

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Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

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