Question

A radioactive sample has $8.0\times {10}^{18}$ active nuclei at a certain instant. How many of these nuclei will still be in the active state after two half-lives (in $\times {10}^{18})$?

Answer 1

Given: $N_0=8\times {10}^{18}$
After one half time, number of nuclei $N=\frac{N_0}{2}=4\times {10}^{18}$
After two half time, number of nuclei $=N/2=1/2\times 4\times {10}^{18}=2\times {10}^{18}$