Question

Rutherfordium-261 has a half-life of 1.08 min. How long will it take for a sample of rutherfordium to lose one-third of its nuclei?

Answer 1

Answer:

$t=1.712min$

Explanation:

Hello!

In this case, since the radioactive decay equation is:

$\frac{A}{A_0}=2^{-\frac{t}{t_{1/2} }$

Whereas A stands for the remaining amount of this sample and A0 the initial one. In such a way, since the sample of rutherfordium is reduced to one-third of its nuclei, the following relationship is used:

$A=\frac{1}{3} A_0$

And we plug it in to get:

$\frac{\frac{1}{3} A_0}{A_0}=2^{-\frac{t}{t_{1/2}} } \\\\\frac{1}{3}=2^{-\frac{t}{t_{1/2}} }$

Now, as we know its half-life, we can compute the elapsed time for such loss:

$log(\frac{1}{3})=log(2^{-\frac{t}{t_{1/2}} })\\\\log(\frac{1}{3})=-\frac{t}{t_{1/2}} }*log(2)$

$t=-\frac{log(\frac{1}{3})t_{1/2}}{log(2)} \\\\t=1.71min$

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