Question

If the half-life of a unstable isotope is 10,00 years and only 1/8 of the radioactive parent remains how old is the sample? A.10,000 B. 20,000 C. 30,000 D. 40,000

Answer 1

IF the half-life of an unstable isotope is 10,000 years and only 1/8 of the radioactive parent remains, the sample is 30,000 years old.

Answer 2

C. 30,000

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 (1)/(2)=1\cdot e^(r\cdot 10000)`

`\Rightarrow (1)/(2)=e^(r\cdot 10000)`

`\Rightarrow \ln (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=(-\ln 2)/(10000)`

Now the function becomes,

`y=ae^{(-\ln 2)/(10000)\cdot t}`

Now, only 1/8 of the radioactive parent remains, so

`\Rightarrow (1)/(8)=1\cdot e^{(-\ln 2)/(10000)\cdot t}`

`\Rightarrow (1)/(8)=e^{(-\ln 2)/(10000)\cdot t}`

`\Rightarrow \ln (1)/(8)=\ln e^{(-\ln 2)/(10000)\cdot t}`

`\Rightarrow -\ln 8={(-\ln 2)/(10000)\cdot t}\cdot \ln e`

`\Rightarrow -\ln 8={(-\ln 2)/(10000)\cdot t}`

`\Rightarrow t=(-\ln 8\cdot 10000)/(-\ln 2)`

`\Rightarrow t=30,000`