Decay constants of two radioactive samples A | vedclass

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(A) The number of nuclei remaining at time $t$ is given by the law of radioactive decay: $N(t) = N_0 e^{-\lambda t}$.For sample $A$,the decay constant

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$e^{-2}$

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(A) The number of nuclei remaining at time $t$ is given by the law of radioactive decay: $N(t) = N_0 e^{-\lambda t}$.For sample $A$,the decay constant $\lambda_A = 15x$. The number of nuclei remaining after time $t = \frac{1}{6x}$ is:$N_A = N_0 e^{-(15x)(\frac{1}{6x})} = N_0 e^{-15/6} = N_0 e^{-5/2}$.For sample $B$,the decay constant $\lambda_B = 3x$. The number of nuclei remaining after time $t = \frac{1}{6x}$ is:$N_B = N_0 e^{-(3x)(\frac{1}{6x})} = N_0 e^{-3/6} = N_0 e^{-1/2}$.The ratio of the number of nuclei left in $A$ and $B$ is:$\frac{N_A}{N_B} = \frac{N_0 e^{-5/2}}{N_0 e^{-1/2}} = e^{-5/2 - (-1/2)} = e^{-4/2} = e^{-2}$.

Decay constants of two radioactive samples $A$ and $B$ are $15x$ and $3x$ respectively. They have an equal number of initial nuclei. The ratio of the number of nuclei left in $A$ and $B$ after a time $t = \frac{1}{6x}$ is

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