Answer:
[tex]t=1.712min[/tex]
Explanation:
Hello!
In this case, since the radioactive decay equation is:
[tex]\frac{A}{A_0}=2^{-\frac{t}{t_{1/2} }[/tex]
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:
[tex]A=\frac{1}{3} A_0[/tex]
And we plug it in to get:
[tex]\frac{\frac{1}{3} A_0}{A_0}=2^{-\frac{t}{t_{1/2}} } \\\\\frac{1}{3}=2^{-\frac{t}{t_{1/2}} }[/tex]
Now, as we know its half-life, we can compute the elapsed time for such loss:
[tex]log(\frac{1}{3})=log(2^{-\frac{t}{t_{1/2}} })\\\\log(\frac{1}{3})=-\frac{t}{t_{1/2}} }*log(2)[/tex]
[tex]t=-\frac{log(\frac{1}{3})t_{1/2}}{log(2)} \\\\t=1.71min[/tex]
Best regards!
