Respuesta :
Answer:
a) [tex]1.065\times 10^{-8} [/tex]fraction of carbon−14 in a piece of charcoal remains after 14.0 years.
b) [tex]0.000\times 10^{-3} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.900\times 10^4 years[/tex]
c) [tex]0.0000\times 10^{-4} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.0000\times 10^5 years[/tex].
Explanation:
The fraction of a radioactive isotope remaining at time t is given by:
[tex][A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}[/tex]
Taking log both sides:
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[A] = fraction at given time t
[tex]t_{\frac{1}{2}}[/tex] = half life of the carbon−14 =5,730 years
a)When , t = 14 years
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 14 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=1.065\times 10^{-8} [/tex]
b)When , t = [tex]1.900\times 10^4 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.000\times 10^{-3} [/tex [/tex]
c)When , t = [tex]1.0000\times 10^5 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.0000\times 10^{-4} [/tex]
