Answer:
C. 30,000
Step-by-step explanation:
We know that, the exponential function for decay is,
\(y=ae^{rt}\)
where,
y = the amount after time t,
a = initial amount,
r = rate of decay.
The half-life of a unstable isotope is 10,000 years, so
\(\Rightarrow \dfrac{1}{2}=1\cdot e^{r\cdot 10000}\)
\(\Rightarrow \dfrac{1}{2}=e^{r\cdot 10000}\)
\(\Rightarrow \ln \dfrac{1}{2}=\ln e^{r\cdot 10000}\)
\(\Rightarrow -\ln 2={r\cdot 10000}\cdot \ln e\)
\(\Rightarrow -\ln 2={r\cdot 10000}\cdot 1\)
\(\Rightarrow r=\dfrac{-\ln 2}{10000}\)
Now the function becomes,
\(y=ae^{\frac{-\ln 2}{10000}\cdot t}\)
Now, only 1/8 of the radioactive parent remains, so
\(\Rightarrow \dfrac{1}{8}=1\cdot e^{\frac{-\ln 2}{10000}\cdot t}\)
\(\Rightarrow \dfrac{1}{8}=e^{\frac{-\ln 2}{10000}\cdot t}\)
\(\Rightarrow \ln \dfrac{1}{8}=\ln e^{\frac{-\ln 2}{10000}\cdot t}\)
\(\Rightarrow -\ln 8={\frac{-\ln 2}{10000}\cdot t}\cdot \ln e\)
\(\Rightarrow -\ln 8={\frac{-\ln 2}{10000}\cdot t}\)
\(\Rightarrow t=\dfrac{-\ln 8\cdot 10000}{-\ln 2}\)
\(\Rightarrow t=30,000\)
