-1
$\begingroup$

Here A is the parent atom and B is tha daughter atom enter image description here

$\endgroup$
  • $\begingroup$ Could you explain what you thought about this process? $\endgroup$ – KV18 Apr 13 '19 at 18:11
1
$\begingroup$

This is an example of the parent nuclei $A$ decaying to produce daughter nuclei $B$ which in turn decay into granddaughter nuclei $C$.

Activity is the rate of decay of unstable nuclei and is proportion to the number of unstable nuclei.

In this example of such a decay there are initially no daughter nuclei (at the start the activity of daughter nuclei is zero) and the half life of the parent nuclei is smaller than the half life of the daughter nuclei.

Initially the rate of decay of parent nuclei is greater than that of the rate at which daughter nuclei decay (there are not many daughter nuclei and their half life is larger that that of the parent nuclei) so the number of daughter nuclei increases with time (activity of daughter nuclei is increasing).

As the number of parent nuclei decreases so the rate of decay of these nuclei decreases whilst at the same time as the number of daughter nuclei is increasing the rate of decay of these nuclei increases.

Eventually there comes at time when the rate at which both parent and daughter are decaying is the same and this is the peak of graph $B$.

After that time the rate of decay of the daughter nuclei is greater than the rate at which they are produced (rate of decay of parent nuclei) and so as time progresses there is a reduction in the number of daughter nuclei and the number of parent nuclei is also decreasing towards zero.


Update

The answer to this question Proving that the water leaving a vertical pipe [with a hole at the bottom] is exponential (decay) can be used as the basis of a water analogy of the parent to daughter to granddaughter decay graphs.
In the answer is is shown that the flow of water from a container with a hole/tube at the end is (approximately) exponential so consider a set up like this.

enter image description here

Two containers (often burettes in a laboratory experiment) one $A$ with a large "hole" of area $a$ and the other $B$ with a small "hole" of area $b$, with $a>b$, start off at time $t=0$ with water in container $A$ and no water in container $B$.

Whilst reading the next passage think of the volume of water as the number of unstable nuclei and the rate of flow of water as the rate at which nuclei are decaying.
The water in the container with the bigger hole $A$, and all other things being equal having the greatest rate of flow of water out of the container, represents the nuclei with the shorter half life.

Water is then allowed to flow out of container $A$ into container $B$ which in turn will have water flowing out of it.

Initially water flows out of container $A$ into container $B$ at a faster rate than water flows out of container $B$ so the water level in container $A$ will go down and the water level in container $B$ will go up.
Number of nuclei type $A$ (or activity of $A$) is decreasing whilst the umber of nuclei type $B$ (or activity of $B$) is increasing.

However as the water level in container $A$ drops so does the rate of flow of water out of the container whereas as the level of water in container $B$ increases so does the rate of flow water out of the container.

With one rate of flow going down and the other rate of flow going up there will come a time when both flow rates are the same.
Peak of the activity of $B$ graph.

After that time the rate at which water flows from container $A$ into container $B$ will be less than the rate at which water is flowing out of container $B$ and the level of water will drop in both containers.
Activity graphs after the peak in the activity of $B$.

$\endgroup$
0
$\begingroup$

I answer by gut feeling.

The daughter atom is generated by the decay of the "parent atom".

When there are many parent atoms, per unit time more daughter atoms are generated than decay (the decay rate is proportional to the number of daughter atoms and the generation rate to the number of parent atoms). This means that the number of daughter atoms increases and therefore the rate of decay (and with it the radioactivity) increases.

As time passes, the number of parent atoms decreases, less daughter atoms are generated and at a certain time their number will start to decrease: from this moment on the radioactivity will decrease.

The shape of the curves is calculated with rather straightforward linear differential equations which are called "rate equations".

$\endgroup$
0
$\begingroup$

By definition, activity $A$ and $B$ are defined through \begin{align} {A_A}\left( t \right) &= \frac{{\# {\text{decayed A atom}}}}{{{\text{unit time}}}} = {\lambda _A}{N_A}\left( t \right) \\ {A_B}\left( t \right) &= \frac{{\# {\text{decayed B atom}}}}{{{\text{unit time}}}} = {\lambda _B}{N_B}\left( t \right). \end{align}

We know that \begin{gather} {{\dot N}_A}\left( t \right) = - {A_A}\left( t \right) = - {\lambda _A}{N_A}\left( t \right) \Rightarrow {N_A}\left( t \right) = {N_A}\left( 0 \right) \cdot {e^{ - {\lambda _A}t}} \end{gather}

therefore \begin{align} \label{eq:de} \tag{1} {{\dot N}_B} & \left( t \right) = \underbrace {\frac{{\# {\text{atom decayed from A to B}}}}{{{\text{unit time}}}}}_{ = r\frac{{\# {\text{decayed A atom}}}}{{{\text{unit time}}}}} - \frac{{\# {\text{decayed B atom}}}}{{{\text{unit time}}}} \\ & = r{\lambda _A}{N_A}\left( t \right) - {\lambda _B}{N_B}\left( t \right) \\ & = \underbrace {r{\lambda _A}{N_A}\left( 0 \right) \cdot {e^{ - {\lambda _A}t}}}_{\alpha \left( t \right)} - \underbrace {{\lambda _B}{N_B}\left( t \right)}_{\beta \left( t \right)}, \\ \end{align} where $r$ is the ratio describing how many atom of type $B$ is created by the decay of one $A$ atom (e.g., it can be considered $1$ for the sake of simplicity).


Method 1

Instead of trying to solve the differential equation \eqref{eq:de}, try to imagine the result. ${\alpha \left( t \right)}$ is positive at $t=0$, while ${\beta \left( t \right)} = 0$. After long enough time, ${\alpha \left( t \right)}=0$, but if ${{\dot N}_B}$ weren't $0$, according to ${{\dot N}_B} = - {\lambda _B}{N_B}\left( t \right)$, it would decay to 0 exponentially. With other words, ${{\dot N}_B}\left( t \right) = 0 \Leftrightarrow {N_B}\left( t \right) = 0$.

So ${N_B}\left( t \right)$ increases in the beginning, starting from $0$, but in the end it doesn't even matter it is $0$ again. If we believe that the funtion is continuous, it must have a maximum.


Method 2

Solve the differential equation \eqref{eq:de} with $r=1$! The general solution is \begin{equation} {N_B}\left( t \right) = {e^{ - {\lambda _B}t}}\left( {C + \frac{{{e^{\left( {{\lambda _B} - {\lambda _A}} \right)t}}}}{{{\lambda _B} - {\lambda _A}}}{\lambda _A}{N_A}\left( 0 \right)} \right). \end{equation} Using the initial condition ${N_B}\left( 0 \right) = 0$, \begin{equation} C + \frac{1}{{{\lambda _B} - {\lambda _A}}}{\lambda _A}{N_A}\left( 0 \right) = 0 \Rightarrow C = - \frac{1}{{{\lambda _B} - {\lambda _A}}}{\lambda _A}{N_A}\left( 0 \right). \end{equation}

Our solution is therefore \begin{align} {N_B}\left( t \right) & = {e^{ - {\lambda _B}t}}\left( {\frac{{{e^{\left( {{\lambda _B} - {\lambda _A}} \right)t}}}}{{{\lambda _B} - {\lambda _A}}}{\lambda _A}{N_A}\left( 0 \right) - \frac{1}{{{\lambda _B} - {\lambda _A}}}{\lambda _A}{N_A}\left( 0 \right)} \right) \\ & = {e^{ - {\lambda _B}t}}\frac{{{\lambda _A}{N_A}\left( 0 \right)}}{{{\lambda _B} - {\lambda _A}}}\left( {{e^{\left( {{\lambda _B} - {\lambda _A}} \right)t}} - 1} \right). \\ \end{align}

Now let's plot the result with $\lambda_A = 2\cdot \lambda_B$.enter image description here

Special case: $\lambda_A = \lambda_B$

In the case $\lambda_A = \lambda_B$, one gets a $0/0$ type limit in the general solution. By executing it one gets $${N_B}\left( t \right) = {e^{ - {\lambda _A}t}}\left( {C + {\lambda _A}{N_A}\left( 0 \right) \cdot t} \right)$$ for the general solution. The initial condition says: $${N_B}\left( 0 \right) = 0 \Rightarrow C = 0,$$ therefore our solution is $${N_B}\left( t \right) = {\lambda _A}{N_A}\left( 0 \right) \cdot t{e^{ - {\lambda _A}t}}.$$

It looks like the following: enter image description here


Interpretation

Imagine the following: there are tons of $A$ atoms decaying, $B$ atoms are generated rapidly. There are only a few $B$ atoms yet, only couple of them decays in the beginning, because there are essentially no $B$ atoms. But the number of $A$ atoms are shrinking, less $B$ atoms are generated, furthermore, because the amount of $B$ atoms are higher, larger number of $B$ atoms decay. There is an equilibrium, when $r{\lambda _A}{N_A}\left( 0 \right) \cdot {e^{ - {\lambda _A}t}} = {\lambda _B}{N_B}\left( t \right)$, therefore ${{\dot N}_B}\left( t \right) = 0 \Rightarrow {{\dot A}_B}\left( t \right) = 0$. In the next moment less $B$ atoms are generated by the decay of $A$ atoms, because there will be even less $A$ atoms, but the decay of $B$ atoms doesn't stop, so the number of $B$ atoms start decreasing.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.