Question

Radioactive material 'A' has a decay constant '8λ' and material 'B' has decay constant 'λ'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be e?

Answer 1

Radioactive decay:

1. It is the term for the spontaneous disintegration of an atomic nucleus in a radioactive substance that causes the emission of radiation from the nucleus.
2. A parent nuclide is one that undergoes radioactive decay, and a daughter nuclide is one that is created during the radioactive process.
3. The Radioactive Formula is given by: $\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda T}}$

Where,

1. N₀ = the initial quantity of the substance and N is the quantity still remaining and not yet decayed.
2. T is the half-life of the decaying quantity
3. e is the Euler’s number equal to 2.71828

$\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda T}}$

Where,

1. λ is the decay constant
2. N_o is the initial number of nuclei

Initially, the number of nuclei is the same so,

${\mathrm{N}}_{\mathrm{A}0}={\mathrm{N}}_{\mathrm{B}0}={\mathrm{N}}_0$

and

${\mathrm{N}}_{\mathrm{B}}/{\mathrm{N}}_{\mathrm{A}}=\mathrm{e}$

For A, ${\mathrm{N}}_{\mathrm{A}}={\mathrm{N}}_0{\mathrm{e}}^{(-8\mathrm{\lambda })\mathrm{t}}$ (1)
For B, ${\mathrm{N}}_{\mathrm{B}}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$ (2)

Dividing equation (2) by equation (1)

$$\begin{gathered} {\mathrm{N}}_{\mathrm{B}}/{\mathrm{N}}_{\mathrm{A}}=\mathrm{e} \\ {\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}/{\mathrm{N}}_0{\mathrm{e}}^{(-8\mathrm{\lambda t})}=\mathrm{e} \\ {\mathrm{e}}^{7\mathrm{\lambda t}}=\mathrm{e} \\ 7\mathrm{\lambda t}=1 \\ \mathrm{t}=1/7\mathrm{\lambda } \end{gathered}$$