91Ó°ÊÓ

1. At, t = 0 the decay rate of radionuclides A and B is given by $R_A=2R_B$.
2. The disintegration constants,${\mathrm{λ}}_A>{\mathrm{λ}}_B$

The population decays exponentially at a rate that depends on the decay constant. And the decay rate depends inversely proportional to the half-life of the substance defines that the number of atoms of the nuclide has decayed by half in number.

The disintegration constant is as follows:

$\mathrm{λ}=\frac{\ln 2}{T_{{\displaystyle \frac{1}{2}}}}$ …… (i)

Here,$T_{\frac{1}{2}}$is the half-life of the substance.

The rate of undecayed sample after a given time is as follows:

$A=A_0{e}^{-\mathrm{λ}t}$ …… (iii)

Given that${\mathrm{λ}}_A={\mathrm{λ}}_B$

So, from equation (i), one can say that$T_{1/2_B}>T_{1/2_A}$

Now, from the given data$R_A=2R_B\mathrm{at}t=0$at, we can say that the initial rate of both the nuclides can be given using equation (ii) as:

Now, let’s us now check if there is another point at which their decay rates is same. Thus, using equation (ii), we can write that

$$\begin{gathered} R_A\text{'}=R_B\text{'} \\ {R}_{oA}{e}^{-{\mathrm{λ}}_{A}t}=R_{oB}{e}^{-{\mathrm{λ}}_{B}t} \\ 2R_{0B}{e}^{-{\mathrm{λ}}_{A}t}=R_{oB}{e}^{-{\mathrm{λ}}_{B}t}(âˆ\mu R_{0A}=2R_{0B}) \\ {e}^{-({\mathrm{λ}}_A-{\mathrm{λ}}_B)}=\frac{1}{2} \end{gathered}$$

Simplify further as:

$$\begin{gathered} \left({\mathrm{λ}}_A-{\mathrm{λ}}_B\right)t=\ln 2 \\ t=\frac{\ln 2}{\left({\mathrm{λ}}_A-{\mathrm{λ}}_B\right)} \end{gathered}$$

Now, the time value is positive as ${\mathrm{λ}}_A>{\mathrm{λ}}_B$, thus there do exist another point at which they have same decay rate simultaneously.