Question 437: Atoms & Nuclei

Question

Two radioactive substances A and B have decay constants ' $5 \lambda$ ' and ' $\lambda$ ' respectively. At $\mathrm{t}=0$, they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $\left(\frac{1}{\mathrm{e}}\right)^2$ after a time interval

Accepted Answer

Number of nuclei remained after time $t$ can be written as $\mathrm{N} = \mathrm{N}_0 \mathrm{e}^{-\lambda t}$
$$\mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}}\quad	ext{.... (i)}$$
and $\mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}\quad	ext{.... (ii)}$
Dividing equation (i) by equation (ii), we get,
$$\begin{aligned}
& \frac{N_1}{N_2}=e^{(-5 \lambda+\lambda) t}=e^{-4 \lambda t}=\frac{1}{e^{4 \lambda \lambda}} \
& \frac{N_1}{N_2}=\left(\frac{1}{e}ight)^2=\frac{1}{e^2} \quad	ext{... [Given]}\
	herefore \quad & \frac{1}{e^2}=\frac{1}{e^{4 \lambda \lambda}} \
	herefore \quad & 2=4 \lambda t \Rightarrow t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}
\end{aligned}$$

Answer

$\frac{1}{42}$

Answer

$4 \lambda$

Answer

$2 \lambda$

Answer

$\frac{1}{2 \lambda}$

Atoms & Nuclei

Two radioactive substances A and B have decay constants ' 5 \lambda ' and ' \lambda ' respectively. At \mathrm{t}=0, they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be \left(\frac{1}{\mathrm{e}}\right)^2 after a time interval

Choose the correct answer: