(a) The daughter nucleus formed in radioactive decay is...

(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. 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 $$ N_{2}(t)=\frac{N_{10} \lambda_{1}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t}-e^{-\lambda_{1}}\right) $$ This equation is the law of successive radioactive decays. (c) $^{218}$ Po decays into $^{214} \mathrm{Pb}$ 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 $\mathrm{Po}$ and $N_{0}(t)$ for 214 Pb. Let $N_{10}=1000$ nuclei and choose values of $t$ from 0 to 36 $\mathrm{min}$ in 2 -min intervals. The curve for 214 $\mathrm{Pb}$ at first rises to a maximum and then starts to decay. At what instant $t_{\mathrm{m}}$ is the number of $^{214 \mathrm{Pb} \text { nuclei a maximum? (d) By }}$ 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 $\lambda_{2}$ . Explain whether the value obtained in part (c) agrees with this equation.

(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. 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
$$
N_{2}(t)=\frac{N_{10} \lambda_{1}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t}-e^{-\lambda_{1}}\right)
$$
This equation is the law of successive radioactive decays. (c) $^{218}$ Po decays into $^{214} \mathrm{Pb}$ 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 $\mathrm{Po}$ and $N_{0}(t)$ for 214 Pb. Let $N_{10}=1000$ nuclei and choose values of $t$ from 0 to 36 $\mathrm{min}$ in 2 -min intervals. The curve for 214 $\mathrm{Pb}$ at first rises to a maximum and then starts to decay. At what instant $t_{\mathrm{m}}$ is the number of $^{214 \mathrm{Pb} \text { nuclei a maximum? (d) By }}$ 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 $\lambda_{2}$ . Explain whether the value obtained in part (c) agrees with this equation.

Key Concepts

Maximum Condition in Decay Chains

When dealing with a radioactive decay chain, the number of daughter nuclei can increase to a maximum value before it starts to decline. The time at which this maximum occurs is found by setting the derivative of the daughter nuclei population with respect to time equal to zero. This condition, dN2/dt = 0, is used to derive an equation linking the decay constants, which is an important step in analyzing and predicting the dynamics of the decay chain.

Successive Radioactive Decay

Successive or chain decay occurs when a parent nucleus decays into a daughter nucleus that is itself radioactive and decays further. This means that the daughter nucleus is both produced by the decay of the parent and lost due to its own decay process. This dual role results in a differential equation with a source term from the parent decay and a decay term that accounts for the daughter’s own instability.

Half-life and Decay Constant Relationship

The half-life of a radioactive substance is the time required for half of the nuclei to decay. This intrinsic property is directly related to the decay constant by the relation t1/2 = ln(2)/?. Understanding this relationship is essential for converting between these two parameters and applying them in decay equations.

Differential Equation Modeling

In the study of radioactive decay, differential equations are used to mathematically describe the rate of change of the number of nuclei. For example, the decay of a substance is represented by a first-order differential equation, which can be extended to account for production and decay processes in successive decays. This approach is critical for solving and predicting the behavior of nuclear systems over time.

Exponential Decay Law

This concept describes how a single radioactive substance diminishes over time, following an exponential decay pattern. It is usually modeled as N(t) = N(0)e^(??t), where N(t) is the number of undecayed nuclei at time t, N(0) is the initial number, and ? is the decay constant. The law is fundamental in understanding how nuclei decrease in number due to radioactive decay.

Transcript

00:01 In this question, we're looking at a sequential decay equation where we have the initial nucleide.

00:11 It's called this 1, nucleic number 1, decay is into nucleic number 2, where nucleide number 2 can still go on to decay further to maybe a stable product.

00:28 So there's 2 decay that's going on over here with two different decay constants.

00:33 We want to find that what is the number rate of change of the number of nucleic number two.

00:46 So nuclear number two, dn2 gains from the decay of nuclear number one.

00:57 So everything that decays from nucleic number one goes to nuclear number two.

01:04 So there's the rate of increase of nuclear number two is equal.

01:09 To lambda 1 times n1 this is the increase in nucleic number 2 but however nucleot number 2 itself also decreases due to the decay into other products right this decay and this decay rate is governed by lambda 2 so lambda 2 as well as the number of nucleide stat is present currently which is n2 right therefore this is our differential equation for our second nucleip.

01:51 And to solve for this is quite a tedious process, but we are given a trial solution.

02:00 That is 2.

02:24 Right, so this is our trial solution, which we want to see whether it actually satisfies our differential equation.

02:34 And we're going to do that is we're going to that into this equation, right? so n2 into the n2 d t as well as into this lambda 2 times n2.

02:49 So first off we find what is the differential, differentiating the trial solution by t.

03:11 So only the exponential terms will be affected.

03:13 The exponential terms will have the lambda terms coming down.

03:33 And the right hand side of the equation where the right hand side is just lambda 1 and 1 minus lambda 2 and 2 right so what we're going to do is you want to bring this lambda 2 and 2 term to the left -hand side so we're going to add lambda 2 and 2 to this equation and the right hand side will be only left with lambda 1 and 1.

04:23 Now we want to see whether this is true, right, whether this agrees with our left -hand side over here.

04:32 And to proceed further, we will need to substitute in n2 again, so this will be lambda 2 times.

04:40 The trial solution, which is given n10 lambda 1 over 1 lambda 2.

05:01 So this term is this now.

05:06 And now we see that we can actually factorize out the common factor, which is this.

05:13 Factor over here.

05:22 And the remaining term for this term is negative lambda 2, e negative.

05:34 But for the second term, the remaining one is this times lambda 2.

06:11 Alright, and so we see that this actually cancels out.

06:14 And what we're left with is n .0 .0 .0 .1 plus by nbda 1 minus lambda 2 times exponential negative lambda 1 t.

06:34 Now we see a very familiar expression over here...

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