# Find ratio of number of atoms of $A$ and $B$

However I am thinking of this question to be damn easy but still I'll not say as I am not getting it,

There are two radio nuclei $A$ and $B$. $A$ is $\alpha$ emitter and $B$ is $\beta$ emitter and their decay constant are in ratio $1:2$. What should be the number of atoms of $A$ and $B$ at time $t=0$, so that probability of getting $\alpha$ and $\beta$ particles are same at $t=0$.

So here how can the decay constant be dependent on number of atoms of $A$ and $B$ so that $\alpha$ and $\beta$ particle are in same ratio? I really need a right approach for this one.

There are two radio nuclei A and B. A is α emitter and B is β emitter and their decay constant are in ratio 1:2

So B decays twice as fast as A. If we want the actual number of decay events from A to equal the actual number of decay events from B at t=0, then we must have twice as much A present as B at t=0.

We are given that $\ce{k_{B} = 2k_{A}}$. For the decay events at T=0 to be equal for A and B, then $$\ce{[B]_{o} * k_{B}~=~[A]_{o} * k_{A}}$$ $$\ce{[B]_{o} * 2k_{A}~=~[A]_{o} * k_{A}}$$ or $$\ce{\frac{2}{1}~=~\frac{[A]_{o}}{[B]_{o}}}$$ At t=0, the concentration of A must be twice the concentration of B in order for equal decay events to occur from A and B