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(A) The number of undecayed nuclei is given by $N = N_0 (1/2)^n$,where $n = t/T_{1/2}$ is the number of half-lives.Given: $t = 2 	ext{ days} = 48 	e

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$1:1$

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(A) The number of undecayed nuclei is given by $N = N_0 (1/2)^n$,where $n = t/T_{1/2}$ is the number of half-lives.Given: $t = 2 	ext{ days} = 48 	ext{ hours}$.For the first element: $n_1 = 48/12 = 4$.For the second element: $n_2 = 48/16 = 3$.The initial ratio is $(N_0)_1 / (N_0)_2 = 2/1$.The ratio of undecayed parts after $48 	ext{ hours}$ is:$\frac{N_1}{N_2} = \frac{(N_0)_1}{(N_0)_2} 	imes \frac{(1/2)^{n_1}}{(1/2)^{n_2}}$$\frac{N_1}{N_2} = \frac{2}{1} 	imes \frac{(1/2)^4}{(1/2)^3} = 2 	imes (1/2)^{4-3} = 2 	imes (1/2)^1 = 1$.Thus,the ratio is $1:1$.

At a certain time,radioactive elements are taken in the ratio $2:1$. Their half-lives are $12$ hours and $16$ hours respectively. What will be the ratio of the undecayed parts after $2$ days?

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