A radioactive nucleus $n_2$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If initial number of both nuclei are the same, what is the ratio of number of nuclei of $n_2$ to the number of nuclei of $n_1$, after one half-life of $n_1$ ?

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] After $t=\frac{\ln2}{\lambda_1}$: $N_1=\frac{N_0}2$ and $N_2=N_0e^{-3\lambda_1t}=N_0e^{-3\ln2}=\frac{N_0}8$. Ratio $N_2:N_1=\frac{1/8}{1/2}=\frac14$. Answer (4).

Q. A radioactive nucleus $n_2$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If initial number of both nuclei are the same, what is the ratio of number of nuclei of $n_2$ to the number of nuclei of $n_1$, after one half-life of $n_1$ ?

Explanation

Reviewed by Dr. Abubacker

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