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

`t=1.712min`

Step-by-step explanation:

Hello!

In this case, since the radioactive decay equation is:

`(A)/(A_0)=2^{-(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=(1)/(3) A_0`

And we plug it in to get:

`((1)/(3) A_0)/(A_0)=2^{-(t)/(t_(1/2)) } \\\\(1)/(3)=2^{-(t)/(t_(1/2)) }`

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

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

`t=-(log((1)/(3))t_(1/2))/(log(2)) \\\\t=1.71min`

Best regards!