Respuesta :
ANSWER
[tex]\begin{equation*} 1200\text{ seconds }=20\text{ minutes} \end{equation*}[/tex]EXPLANATION
To calculate the half-life of the sample, apply the formula for the quantity remaining in a radioactive decay:
[tex]\frac{N}{N_o}=(\frac{1}{2})^n[/tex]where N/No = ratio of quantity remaining to initial quantity = 1/16
and n is:
[tex]n=\frac{T}{t_{\frac{1}{2}}}[/tex]where T = time elapsed = 80 mins = 4800 seconds
t1/2 = half-life
Solving for n in the equation above:
[tex]\begin{gathered} \frac{1}{16}=(\frac{1}{2})^n \\ \\ (\frac{1}{2})^4=(\frac{1}{2})^n \\ \\ \Rightarrow n=4 \end{gathered}[/tex]Therefore, the half-life of the sample is:
[tex]\begin{gathered} 4=\frac{4800}{t_{\frac{1}{2}}} \\ \\ t_{\frac{1}{2}}=\frac{4800}{4} \\ \\ t_{\frac{1}{2}}=1200\text{ seconds }=20\text{ minutes} \end{gathered}[/tex]That is the answer.
