Question

(a) The daughter nucleus formed in radioactive decay is often radioactive. Let $N_{10}$ represent the number of parent nuclei at time $t=0, N_{1}(t)$ the number of parent nuclei at time $t,$ and $\lambda_{1}$ the decay constant of the parent. Suppose the number of daughter nuclei at time $t=0$ is zero. Let $N_{2}(t)$ be the number of daughter nuclei at time $t$ and let $\lambda_{2}$ be the decay constant of the daughter. Show that $N_{2}(t)$ satisfies the differential equation $$\frac{d N_{2}}{d t}=\lambda_{1} N_{1}-\lambda_{2} N_{2}$$ (b) Verify by substitution that this differential equation has the solution $$N_{2}(t)=\frac{N_{10} \lambda_{1}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t}-e^{-\lambda_{1} t}\right)$$ This equation is the law of successive radioactive decays. (c) 218 Po decays into 214 Pb with a half-life of 3.10 $\mathrm{min}$ , and 214 Pb decays into 214 Bi with a half-life of 26.8 min. On the same axes, plot graphs of $N_{1}(t)$ for $^{218}$ Po and $N_{2}(t)$ for 214 Pb. Let $N_{10}=1000$ nuclei and choose values of $t$ from 0 to 36 $\mathrm{min}$ in 2 -min intervals. (d) The curve for 219 $\mathrm{Pb}$ obtained in part $(\mathrm{c})$ at first rises to a maximum and then starts to decay. At what instant $t_{m}$ is the number of 214 $\mathrm{Pb}$ nuclei a maximum? (e) By applying the condition for a maximum $d N_{2} / d t=0,$ derive a symbolic equation for $t_{m}$ in terms of $\lambda_{1}$ and $\lambda_{2} .$ (f) Explain whether the value obtained in part (c) agrees with this equation.

Name: Problem 32, Chapter 44: Physics for Scientists and Engineers with Modern Physics
Uploaded: 2020-09-13T07:44:24-08:00
Duration: 19 min 38 s
Channel: Ren Jie Tuieng

   (a) The daughter nucleus formed in radioactive decay is often radioactive. Let $N_{10}$ represent the number of parent nuclei at time $t=0, N_{1}(t)$ the number of parent nuclei at time $t,$ and $\lambda_{1}$ the decay constant of the parent. Suppose the number of daughter nuclei at time $t=0$ is
zero. Let $N_{2}(t)$ be the number of daughter nuclei at time $t$ and let $\lambda_{2}$ be the decay constant of the daughter. Show that $N_{2}(t)$ satisfies the differential equation
$$\frac{d N_{2}}{d t}=\lambda_{1} N_{1}-\lambda_{2} N_{2}$$
(b) Verify by substitution that this differential equation has the solution
$$N_{2}(t)=\frac{N_{10} \lambda_{1}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t}-e^{-\lambda_{1} t}\right)$$
This equation is the law of successive radioactive decays. (c) 218 Po decays into 214 Pb with a half-life of 3.10 $\mathrm{min}$ , and 214 Pb decays into 214 Bi with a half-life of 26.8 min. On the
same axes, plot graphs of $N_{1}(t)$ for $^{218}$ Po and $N_{2}(t)$ for 214 Pb. Let $N_{10}=1000$ nuclei and choose values of $t$ from 0 to 36 $\mathrm{min}$ in 2 -min intervals. (d) The curve for 219 $\mathrm{Pb}$ obtained in part $(\mathrm{c})$ at first rises to a maximum and then starts to decay. At what instant $t_{m}$ is the number of 214 $\mathrm{Pb}$ nuclei a maximum? (e) By applying the condition for a maximum $d N_{2} / d t=0,$ derive a symbolic equation for $t_{m}$ in terms of $\lambda_{1}$ and $\lambda_{2} .$ (f) Explain whether the value obtained in part (c) agrees with this equation.

Physics for Scientists and Engineers with Modern Physics

Raymond A. Serway,… 8th Edition

Chapter 44, Problem 32

Instant Answer

Solved by Expert Ren Jie Tuieng

09/13/2020

Step 1

The first is the decay of the parent nuclei, $N_{1}$, which produces daughter nuclei at a rate proportional to the number of parent nuclei present. This is represented by the term $\lambda_{1} N_{1}$. The second process is the decay of the daughter nuclei Show more…

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(a) The daughter nucleus formed in radioactive decay is often radioactive. Let $N_{10}$ represent the number of parent nuclei at time $t=0, N_{1}(t)$ the number of parent nuclei at time $t,$ and $\lambda_{1}$ the decay constant of the parent. Suppose the number of daughter nuclei at time $t=0$ is zero. Let $N_{2}(t)$ be the number of daughter nuclei at time $t$ and let $\lambda_{2}$ be the decay constant of the daughter. Show that $N_{2}(t)$ satisfies the differential equation $$\frac{d N_{2}}{d t}=\lambda_{1} N_{1}-\lambda_{2} N_{2}$$ (b) Verify by substitution that this differential equation has the solution $$N_{2}(t)=\frac{N_{10} \lambda_{1}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t}-e^{-\lambda_{1} t}\right)$$ This equation is the law of successive radioactive decays. (c) 218 Po decays into 214 Pb with a half-life of 3.10 $\mathrm{min}$ , and 214 Pb decays into 214 Bi with a half-life of 26.8 min. On the same axes, plot graphs of $N_{1}(t)$ for $^{218}$ Po and $N_{2}(t)$ for 214 Pb. Let $N_{10}=1000$ nuclei and choose values of $t$ from 0 to 36 $\mathrm{min}$ in 2 -min intervals. (d) The curve for 219 $\mathrm{Pb}$ obtained in part $(\mathrm{c})$ at first rises to a maximum and then starts to decay. At what instant $t_{m}$ is the number of 214 $\mathrm{Pb}$ nuclei a maximum? (e) By applying the condition for a maximum $d N_{2} / d t=0,$ derive a symbolic equation for $t_{m}$ in terms of $\lambda_{1}$ and $\lambda_{2} .$ (f) Explain whether the value obtained in part (c) agrees with this equation.

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Key Concepts

Maximum Condition in Decay Processes

When modeling the population of a daughter nucleus, there can be a point where its number reaches a maximum before eventually decaying away. Determining this maximum involves setting the derivative of the daughter nucleus concentration with respect to time equal to zero. This mathematical condition is used to derive symbolic expressions that pinpoint the time at which the maximum concentration occurs.

Half-Life

The half-life of a radioactive substance is the time required for half of the original nuclei to decay. It is intrinsically linked to the decay constant and provides an intuitive measure of a substance's stability. In problems involving successive decays, comparing the half-lives of the parent and daughter nuclei helps predict the evolution of the system over time.

Successive Radioactive Decay

Successive radioactive decay occurs when a parent nucleus decays into a daughter nucleus that is itself unstable and subject to further decay. This concept involves coupled differential equations, where the rate of change of the daughter nucleus population depends on both its own decay and its production from the decay of the parent. It is essential for understanding complex decay chains.

Decay Constant

The decay constant, denoted by ?, is a measure of the probability per unit time that a given nucleus will decay. It directly relates to the exponential decay law and provides a link between the rate of decay and measurable quantities such as half-life. Understanding the decay constant is crucial for modeling both single and successive radioactive decays.

Radioactive Decay Law

This fundamental principle describes how the quantity of a radioactive substance decreases over time. It is expressed mathematically by an exponential decay function involving a decay constant, reflecting the probability per unit time that a nucleus will decay. The law is key in calculating the number of parent nuclei remaining over time and forms the basis for more complex decay processes.

Differential Equations

Differential equations are used to mathematically model the rate of change of nuclear populations in radioactive decay. In decay chains, these equations account for the rate at which parent nuclei decay and the formation (and subsequent decay) of daughter nuclei. They provide a framework to derive solutions that describe the time evolution of each nuclide in the decay process.

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Transcript

00:01 In this question we're looking at a two -step decay in which a parent nucleic, just called n1, decays into a radioactive daughter nucleine, n2, and this further decays into a final stable state.

00:18 So what you want to first find is the equation governing the rate of change of the daughter nucleine.

00:24 So the n2 over the t.

00:28 So there is two different processes that affect the number of daughter nuclei.

00:36 So first is the production of daughter nucleic from n1 and also the decay of the daughter nucleate into the more stable compound.

00:50 So this two governs the rate of change of our daughter nuclei.

00:54 So the production increase in daughter nucleate comes from the decay of the decay of the rate of the rate of change of our daughter nucleic.

00:58 N1 so this must be equals to n lambda 1 times n1 right where lambda 1 is the decay constant associated with this decay and n1 is just the number of perennuclite at that particular point in time so this tells us the number of decays of n1 which also relates to the increase in the daughter nuclei then we subtract away the decay of the n2 itself because it's radioactive so there's some decay constant associated with this decay and this decay, the rate of decay will be proportional to the number of daughter nuclei that you have at that point in time.

01:41 So we have to subtract lambda 2 times n2 because it's a decay.

01:46 So that's a negative sign.

01:50 This is a production, producing n2 from the decay of n1.

01:57 So this is positive.

01:58 So therefore this must satisfy the rate of change of the daughter nucleic.

02:09 Now you want to show by substitution that this differential equation can be solved into equals 2 and 1 0 times lambda 1 over lambda 1 minus lambda 2 times the exponential terms.

02:29 So this n1 node over here is the initial number of parent, duplite.

02:48 It's a constant.

02:52 So to show that it satisfies this equation, we use this equation and we're going to differentiate this.

03:00 We're going to differentiate this, the n2 over d t.

03:07 Different with respect to time.

03:15 So this entire thing over here is a constant.

03:20 So we can keep it on one side then we differentiate the exponential terms inside so we have negative negative lambda 2 times the exponential term all right and with this we want to further simplify this all right so we can add in what we're going to do is we're going to add in negative lambda 2 times exponential negative lambda 1 t then we've got to subtract it away as well so that it doesn't affect the original expression.

05:15 So the reason why we add this in into this bracket over here is because that now we can separate out these two terms into to one bracket.

05:34 So we keep this constants as k.

05:37 I'm going to use k for negative lambda 2 times exponential.

05:54 Then we have the additional term, extra term of lambda 2 exponential negative lambda 1 t minus lambda 2 and then the final one which is plus lambda 1, exponential negative lambda 1 t now this additional term that we added is just 0 right it doesn't affect the expression right but what you can do is we can simplify this by segregating these two terms and these two terms so this two becomes negative lambda 2 times exponential lambda 2 t minus exponential negative lambda 1 t.

07:05 Markplied by k of course and we see that this is our original expression for n2 right our n2 expression is just k times exponential negative lambda 2 t minus negative exponential negative lambda 1 t so this is negative lambda 2 times and 2 what about for the other term the other term over here we have k which are we going to write in full right the k is n1 0 times lambda 1 divided by lambda 1 minus lambda 2 this k multiplied by the whatever that's inside the bracket this is negative lambda 2 plus lambda 1 times exponential negative lambda 1 t so we see that these two brackets are the same term, right? it's just lambda 1 minus lambda 2, which we can get rid of.

08:26 We can just divide them away.

08:31 And we are left with n1 times lambda 1 times exponential negative lambda 1 t.

08:43 N1 times this exponential term gives us n1.

08:49 This is the number of of perianuclite at that point in time and therefore this is just n1 times lambda 1 so in all substituting back this is therefore equals to negative lambda 2 and 2 plus lambda 1n which is basically our initial differential equation and therefore it shows that the the expression over here for n2 satisfies the differential equation and so now we're going to use this expression to solve for a real -life example which is polonium 218 which decays into lead to 144 and this led to 114 further decays into bismuth so we know we can calculate the respective respective decay constants, right lambda 1 and lambda 2 based on the half lives that were given.

10:14 So lambda 1 given that the decay from polonium to lead is a half life of 3 .1 minutes so this 3 .1 minute and for lambda 2 this is 26 .8 minutes.

10:51 We also know that the initial number of nucleates n1 is given as 1000, 1000 nucleates and we want to plot the graph for n1 as well as for n2.

11:15 So for n1 is quite simple.

11:19 For n1 is just n1 not times exponential decay...

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