Question

The decay constant of ${}^{238}Uis1.54\times {10}^{-10}y{r}^{-1}$. What will be the average life of a sample of ${10}^{20}such{}^{238}U$ nuclei?

• A. 3.25 billion years
• B. 6.49 million years
• C. 3.25 million years
• D. 6.49 billion years

Answer 1

The correct option is D 6.49 billion years

We know, the average life of a radioactive sample, $t_{avg}$, is related to the decay constant $\lambda$ as-
$t_{avg}=\frac{1}{\lambda }$
Very simple formula for a rather complicated phenomenon. Knowing $\lambda$, we can calculate,
$t_{avg}=\frac{1}{\lambda }=\frac{1}{1.54\times {10}^{-10}y{r}^{-1}}=6.49\times {10}^{9}yrs\Rightarrow t_{avg}=6.49billionyears$