Question 154: Atoms & Nuclei

Question

A radioactive nucleus $\mathrm{n}_2$ 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 $n_1$ ?

Accepted Answer

Let's work through the problem step by step.
The decay of a radioactive nucleus is given by the formula:
$$N(t) = N_0 \, e^{-\lambda t},$$
where:
$N(t)$ is the number of nuclei remaining at time $t$,
$N_0$ is the initial number of nuclei,
$\lambda$ is the decay constant.
For nucleus $n_1$ with decay constant $\lambda_1$, its half-life $t_{1/2}$ is:
$$t_{1/2} = \frac{\ln 2}{\lambda_1}.$$
After one half-life of $n_1$, the number of $n_1$ nuclei remaining is:
$$N_1 = N_0 \, e^{-\lambda_1 \, t_{1/2}} = N_0 \, e^{-\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-\ln2} = \frac{N_0}{2}.$$
Nucleus $n_2$ has a decay constant that is 3 times that of $n_1$:
$$\lambda_2 = 3\lambda_1.$$
For nucleus $n_2$, the number of nuclei remaining after one half-life of $n_1$ (which is $t = \frac{\ln2}{\lambda_1}$) is:
$    N_2 = N_0 \, e^{-\lambda_2 \, t} = N_0\, e^{-3\lambda_1 \, \frac{\ln2}{\lambda_1}} = N_0 \, e^{-3\ln2} = N_0 \cdot \frac{1}{2^3} = \frac{N_0}{8}.    $
Now, we find the ratio of $n_2$ nuclei to $n_1$ nuclei after this time:
$    \frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1/8}{1/2} = \frac{1}{4}.    $
Therefore, the correct ratio is $\frac{1}{4}$, which corresponds to Option A.

Answer

1/4

Answer

1/8

Answer

4

Answer

8

Atoms & Nuclei

A radioactive nucleus \mathrm{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 ?

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