Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be:

We have two different radioactive nuclides, A and B, each starting with the same initial number of nuclei, say $$N_{0}$$. Their half-lives are given as $$T_{1/2}^{(A)} = 10\ \text{min}$$ and $$T_{1/2}^{(B)} = 20\ \text{min}$$, while the elapsed time is $$t = 60\ \text{min}$$.

For radioactive decay we use the law

$$N = N_{0}\left(\dfrac12\right)^{t/T_{1/2}}$$

where $$N$$ is the number of undecayed nuclei left after time $$t$$ and $$T_{1/2}$$ is the half-life.

First, we find the remaining nuclei of A after 60 minutes.

For A, $$t/T_{1/2}^{(A)} = 60/10 = 6$$, so

$$N_{A} = N_{0}\left(\dfrac12\right)^{6} = N_{0}\left(\dfrac{1}{2^{6}}\right) = N_{0}\left(\dfrac{1}{64}\right) = \dfrac{N_{0}}{64}.$$

Next, we find the remaining nuclei of B after 60 minutes.

For B, $$t/T_{1/2}^{(B)} = 60/20 = 3$$, hence

$$N_{B} = N_{0}\left(\dfrac12\right)^{3} = N_{0}\left(\dfrac{1}{2^{3}}\right) = N_{0}\left(\dfrac{1}{8}\right) = \dfrac{N_{0}}{8}.$$

The number of nuclei that have decayed is the difference between the initial and the remaining numbers.

For A: $$D_{A} = N_{0} - N_{A} = N_{0} - \dfrac{N_{0}}{64} = N_{0}\left(1 - \dfrac{1}{64}\right) = N_{0}\left(\dfrac{64}{64} - \dfrac{1}{64}\right) = N_{0}\left(\dfrac{63}{64}\right).$$

For B: $$D_{B} = N_{0} - N_{B} = N_{0} - \dfrac{N_{0}}{8} = N_{0}\left(1 - \dfrac{1}{8}\right) = N_{0}\left(\dfrac{8}{8} - \dfrac{1}{8}\right) = N_{0}\left(\dfrac{7}{8}\right).$$

Now we form the ratio of the decayed nuclei:

$$\dfrac{D_{A}}{D_{B}} = \dfrac{N_{0}\left(\dfrac{63}{64}\right)} {N_{0}\left(\dfrac{7}{8}\right)} = \dfrac{63}{64}\times\dfrac{8}{7} = \dfrac{63\times 8}{64\times 7} = \dfrac{504}{448} = \dfrac{9}{8}.$$

Thus the ratio of the numbers of decayed nuclei of A to B after 60 minutes is $$9:8$$.

Hence, the correct answer is Option D.