Question

Two radioactive materials A and B have decay constants \[6\lambda\] and \[2\lambda\] respectively. If initially they have same number of nuclei, then the ratio of the number of nuclei of A to that of B will be \[\frac{1}{\mathrm{e}}\] after time ______.

पर्याय

• \[\frac{1}{3\lambda}\]

• \[\frac{1}{2\lambda}\]

• \[\frac{1}{\lambda}\]

• \[\frac{1}{4\lambda}\]

Answer 1

Two radioactive materials A and B have decay constants \[6\lambda\] and \[2\lambda\] respectively. If initially they have same number of nuclei, then the ratio of the number of nuclei of A to that of B will be \[\frac{1}{\mathrm{e}}\] after time \[\frac{1}{4\lambda}\].

Explanation:

Number of nuclei remained after time t can be written as \[\mathrm{N=N_0e^{-\lambda t}}\]

\[\mathrm{N}_1=\mathrm{N}_0\mathrm{e}^{-6\lambda\mathrm{t}}\quad....(\mathrm{i})\]

and \[\mathrm{N}_2=\mathrm{N}_0\mathrm{e}^{-2\lambda\mathrm{t}}\quad....(\mathrm{ii})\]

Dividing equation (i) by equation (ii), we get,

\[\frac{\mathrm{N}_1}{\mathrm{N}_2}=\mathrm{e}^{(-6\lambda+2\lambda)\mathrm{t}}=\mathrm{e}^{-4\lambda\mathrm{t}}=\frac{1}{\mathrm{e}^{4\lambda\mathrm{t}}}\]

\[\frac{\mathrm{N}_1}{\mathrm{N}_2}=\frac{1}{\mathrm{e}}\]          ...[Given]

\[\therefore\quad\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{e}^{4\lambda\mathrm{t}}}\]

\[\therefore\quad1=4\lambda\mathrm{t}\Longrightarrow\mathrm{t}=\frac{1}{4\lambda}\]