Question

Two radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At $t=0$ they have the same number of nuclei. Find the time interval at which the ratio of the number of nuclei of A to those of B will be $\frac{1}{e^2}$.

Answer 1

Step 1: Given data

Two radioactive substances A and B have decay constants $5\lambda$ and $\lambda$.

$\frac{N_A}{N_B}=\frac{1}{e^2}=e^{-2}$

Step 2: Concept or formula used

Radioactivity is the phenomenon of spontaneous disintegration of the atomic nucleus by the emission of highly penetrating radiations. The law of radioactive disintegration states that the rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that instant.

For a radioactive substance number of nuclei (N) at any instant of time (t) is described by the law of radioactive disintegration.

$N=N_0{e}^{-\lambda t}$

Here,

$N$ is the number of nuclei.

$\lambda$ is the decay constant.

$N_0$ is the number of nuclei in the beginning.

Step 3: Calculation

Now the ratio of the number of nuclei for atoms A and B with decay constant $5\lambda$ and $\lambda$ respectively are:

$$\begin{gathered} \frac{N_A}{N_B}=\frac{N_0{e}^{-5\lambda t}}{N_0{e}^{-\lambda t}} \\ =e^{-5\lambda t+\lambda t} \\ =e^{-4\lambda t}...\left(1\right) \end{gathered}$$

Therefore from equation (1), we get

$$\begin{gathered} e^{-4\lambda t}=e^{-2} \\ -4\lambda t=-2 \\ t=\frac{-2}{-4\lambda } \\ =\frac{1}{2\lambda } \end{gathered}$$

The time interval at which the ratio of the number of nuclei of A to those of B will be$\frac{1}{e^2}is\frac{1}{2\lambda }$.