Question

If ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{=}\mathit{\lambda }$in (3.10), then$\mathbf{\int }{\mathit{e}}^{\left({\lambda }_2-{\lambda }_1\right)\mathbf{t}}\mathit{d}\mathit{t}\mathbf{=}\mathbf{\int }\mathit{d}\mathit{t}$. Find${\mathit{N}}_{\mathbf{2}}$for this case.

Short answer

The value of the given expression is,$N_2=N_0\lambda t{e}^{-\lambda t}$ .

Step-by-step solution

Given Information.

The given two equations are,

$$\begin{gathered} \int e^{\left({\lambda }_2-{\lambda }_1\right)t}dt=\int dt \\ {\lambda }_1={\lambda }_2=\lambda \end{gathered}$$.

The formula that is used.

The below equation defines the process of decaying is,

${\mathit{N}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathbf{=}\mathbf{\int }{\mathit{\lambda }}_{\mathbf{1}}{\mathit{N}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathit{d}\mathit{t}\mathbf{+}\mathit{c}$.

Here, ${\mathit{N}}_{\mathbf{0}}\mathbf{=}$number of radium atoms at$\mathit{t}\mathbf{=}\mathbf{0}$ ,

${\mathit{N}}_{\mathbf{1}\mathbf{=}}$Number of radium atoms at time,

${\mathit{N}}_{\mathbf{2}\mathbf{=}}$Number of radon atoms at time,

${\mathit{\lambda }}_{\mathbf{1}}$and ${\mathit{\lambda }}_{\mathbf{2}}$decay constants for${\mathit{R}}_{\mathbf{a}}$ and ${\mathit{R}}_{\mathbf{n}}$.

Find the value of N2.

Substitute the given information in the formula.

$$\begin{gathered} N_2{e}^{\lambda t}=\lambda N_0\int dt \\ =\lambda N_{0}t+C \\ {N}_2=\left(\lambda N_{0}t+C\right)e^{-\lambda t} \end{gathered}$$

Use the given initial condition in order to find the value of integration constant, that is, put $N_2=0,t=0$which gives$C=0$.

Thus, the final answer is,

$N_2=0,t=0$.