Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have an equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

We have two radioactive elements A and B whose half-lives are given as $$T_{1/2,A}=20\ \text{min}$$ and $$T_{1/2,B}=40\ \text{min}$$. At $$t=0$$ the two samples contain the same number of nuclei; let this common initial number be $$N_0$$.

The law of radioactive decay states that the number of undecayed nuclei at any time $$t$$ is

$$N(t)=N_0\left(\dfrac12\right)^{t/T_{1/2}}\;.$$

First we find the number of nuclei of element A that remain after $$t=80\ \text{min}$$. For A the exponent is

$$\dfrac{t}{T_{1/2,A}}=\dfrac{80}{20}=4\;,$$

so

$$N_A(80)=N_0\left(\dfrac12\right)^4=N_0\left(\dfrac1{16}\right)=\dfrac{N_0}{16}\;.$$

The number of A nuclei that have decayed by this time is therefore

$$\text{Decayed}_A = N_0 - N_A(80)=N_0-\dfrac{N_0}{16}=\dfrac{15N_0}{16}\;.$$

Now we perform the same calculation for element B. Here the exponent is

$$\dfrac{t}{T_{1/2,B}}=\dfrac{80}{40}=2\;,$$

giving

$$N_B(80)=N_0\left(\dfrac12\right)^2=N_0\left(\dfrac14\right)=\dfrac{N_0}{4}\;.$$

The number of B nuclei that have decayed in 80 minutes is therefore

$$\text{Decayed}_B = N_0 - N_B(80)=N_0-\dfrac{N_0}{4}=\dfrac{3N_0}{4}\;.$$

We now form the required ratio of the numbers of decayed nuclei:

$$\dfrac{\text{Decayed}_A}{\text{Decayed}_B} =\dfrac{\dfrac{15N_0}{16}}{\dfrac{3N_0}{4}} =\dfrac{15}{16}\times\dfrac{4}{3} =\dfrac{15\times4}{16\times3} =\dfrac{60}{48} =\dfrac54\;.$$

Thus the ratio of the decayed numbers of A to B after 80 minutes is $$5:4$$.

Hence, the correct answer is Option B.