# Chain Decay (A-B-C) (Show My Homework: Nuclear Physics)

There are cases in nature when the nuclei $B$ produced by the disintegration of nuclei $A$ are also radioactive.

a) Find the law of variation with time of of the number of nuclei of type $B$, if the initial sample contains $N_0$ nuclei of type $A$. The (exponential) decay constant of $A$ type nuclei is $\lambda_A$ and of $B$ type nuclei is $\lambda_B$.

b) Find the time $t$ for which the number of $B$ nuclei is maximum?

c) If $C$ type nuclei (those obtained after $B$ nuclei disintegration) are stable, find also the law of variation with time for these nuclei.

The described decay chain is : $A\rightarrow ^{\lambda_A}\rightarrow B\rightarrow ^{\lambda_B}\rightarrow C (stable)$

The number of remaining $A$ nuclei after time $t$ is $N_A(t)=N_0*e^{-\lambda_A*t}$

In the time interval $d t$ a number of $N_A\lambda_A*d t$ nuclei of $A$ type transforms into $B$ type.

In the same interval $d t$ a number of $N_B\lambda_B*d t$ nuclei of type $B$ transforms into $C$ type.

Therefore the variation of the number of $B$ type nuclei is

$d N_B =N_A\lambda_A*d t-N_B\lambda_B*d t$

or written a bit differently:

$(d N_B/d t) +N_B\lambda_B =\lambda_AN_0e^{-\lambda_A*t}$

If one multiplies the above relation with $e^{\lambda_B*t}$ one obtains:

$\frac{d}{d t}[N_B(t)e^{\lambda_B*t}] =\lambda_AN_0e^{-(\lambda_A-\lambda_B)*t}$

By integrating the above

$\int_{0}^{N_B}d[N_B(t)e^{\lambda_B*t}] =\int_{0}^{t}\lambda_AN_0e^{-(\lambda_A-\lambda_B)*t}*d t$

__one obtains the law of variation of $B$ type nuclei:__

$N_B(t) =\frac{\lambda_AN_0}{\lambda_B-\lambda_A}*\left ( e^{-\lambda_A*t}-e^{-\lambda_B*t} \right )$ (1)

b) From the condition $d N_B(t_M)/d t =0$ one obtains

$t_M =\frac {1}{\lambda_B -\lambda_A}*ln (\lambda_B/\lambda_A)$ (2)

c) Now we have

$\frac{d N_C(t)}{d t}=\lambda_B N_B(t)$

where $N_B(t)$ was found above (1)

Solving (integrating again like at point a) one gets:

$N_C(t) =N_0\left ( \frac{\lambda_B}{\lambda_A-\lambda_B}e^{-\lambda_A*t}+ \frac{\lambda_A}{\lambda_B-\lambda_A}e^{-\lambda_B*t}+1 \right )$ (3)