Answer:
0.24 grams of polonium is in the sample 966 days later.
Step-by-step explanation:
Given : Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days.
To find : How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?
Solution :
The half-life of polonium-210 is about 138 days.
We have to find the number of half-lives of polonium-210 occur in 966 days.
The number of half-lives is [tex]n=\frac{966}{138}=7[/tex]
The amount of polonium-210 remaining after t days is given by the equation,
[tex]N(t)=N_0\times (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}[/tex]
where,
N(t) is the amount of substance remaining after t days,
[tex]N_0=31[/tex] is the initial amount of substance,
t=966 is the time in days,
[tex]t_{\frac{1}{2}}=138[/tex] is the half-life of the substance.
Substitute the values in the formula,
[tex]N(t)=31 \times (\frac{1}{2})^{\frac{966}{138}[/tex]
[tex]N(t)=31 \times (\frac{1}{2})^{7}[/tex]
[tex]N(t)=31 \times 0.0078125[/tex]
[tex]N(t)=0.24[/tex]
Therefore, 0.24 grams of polonium is in the sample 966 days later.
