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The concept of radioactive half-life was described in Section 13.4, and Equation \(13.11\) gives the relationship between \(T_{1 / 2}\) and \(\lambda\). Another parameter that is often useful in the description of radioactive processes is the mean life, \(\tau\). Although the half-life of a radioactive isotope is accurately known, it is not possible to predict the time when any individual atom will decay. The mean life is a measure of the average length of existence of all the atoms in a particular sample. Show that \(\tau=1 / \lambda\). (Hint: Remember that \(\tau\) is essentially an average value, and use the fact that the number of atoms that decay between \(t\) and \(t+d t\) is equal to \(d N\). Furthermore, note that these \(d N\) atoms have a finite time of existence, \(t\) )

Short Answer

Expert verified
The mean life \(\tau\) of a radioactive isotope is equal to the inverse of the decay constant \(\lambda\), i.e. \(\tau = 1/\lambda\).

Step by step solution

01

Definition of Mean Life

We first need to understand the concept of the 'mean life' of a radioactive isotope. The mean life, \(\tau\), is defined as the average time an atom will exist before it decays. Mathematically, it can be expressed as \(\tau = \int_0^{\infty} t \cdot f(t) dt\), where \(f(t)\) is the probability density function.
02

Applying Decay Law

The decay law for a radioactive substance can be defined as \(f(t) = \lambda e^{-\lambda t}\). Substituting this into our definition of the mean life we get: \(\tau = \int_0^{\infty} t \cdot \lambda e^{-\lambda t} dt\).
03

Solving the Integral

We now need to solve the above integral. We can use the method of integration by parts, with \(u=t\) and \(dv= \lambda e^{-\lambda t} dt\). Computing this, we find that \(\tau = \int_0^{\infty} t \cdot \lambda e^{-\lambda t} dt = 1/\lambda\). Thus, we have demonstrated that \(\tau = 1/\lambda\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Radioactive Half-Life and Mean Life
The concepts of radioactive half-life and mean life play crucial roles in studying radioactive isotopes and their decay. The half-life (\(T_{1/2}\)) tells us the time required for half of the radioactive atoms in a sample to decay, but it doesn't provide information about when specific atoms will decay. Here is where the concept of mean life, denoted as \(\tau\), becomes significant. Mean life represents the average lifespan of all the atoms before they decay, providing a broader statistical view.

To illustrate, imagine a large group of atoms; each has its clock ticking to an unpredictable decay moment. You cannot tell exactly when each will 'chime,' but you can estimate an average 'chime time' for the entire group—that's the mean life. Now, if we think of the decay of atoms as a natural 'clock,' the mean life is akin to an average time each of these clocks runs before stopping.

The mathematical expression to calculate the mean life is \(\tau = 1/\lambda\), where \(\lambda\) is the decay constant—a rate that quantifies how quickly atoms of a radioactive isotope undergo decay. In essence, the higher the decay constant, the shorter the mean life, indicating a faster decaying substance.
The Decay Law for Radioactive Substances
To grasp the decay law for radioactive substances, one must understand their probabilistic nature. The decay law mathematically describes how the quantity of undecayed atoms decreases over time. It is captured by the equation \(N(t) = N_0e^{-\lambda t}\), where \(N(t)\) is the number of undecayed atoms at time \(t\), \(N_0\) is the initial number, and \(\lambda\) is the decay constant.

For a more in-depth insight, consider the probability density function \(f(t) = \lambda e^{-\lambda t}\). This function represents the likelihood of an atom decaying at time \(t\). The integral of this function over time gives us mean life as it accounts for all possible 'decay times' weighted by their probability of occurrence.

By delving into the interplay between the decay law and the concept of mean life, one comprehends that the inevitable randomness of radioactive decay is not so chaotic after all—it adheres to a predictable pattern over a large sample and extended time, governed by this prolific decay law.
Applying Integration by Parts to Solve for Mean Life
The technique of integration by parts is a powerful tool in calculus, especially when working with products of functions. It serves as the mathematical bridge connecting the decay law and mean life of a radioactive isotope. The formula for integration by parts is given by \(\int u \(\text{d}v\) = uv - \int v \(\text{d}u\)\).

For our problem, we set \(u = t\) and \(dv = \lambda e^{-\lambda t} \(\text{d}t\)\), resulting in the integral \(\int_0^{\infty} t \lambda e^{-\lambda t} \(\text{d}t\)\). Using integration by parts, we find not just an abstract formula but a concrete number that encapsulates the essence of a radioactive sample's behavior over time—the mean life.

Through this computation, we get \(\tau = 1/\lambda\), a surprisingly simple yet profoundly meaningful outcome. This process illuminates how a seemingly challenging calculation can be broken down and decoded using a clever algebraic methodology.

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Most popular questions from this chapter

(a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle with an initial energy of \(0.5 \mathrm{MeV}\) and a gold nucleus \(\left({ }^{197} \mathrm{Au}\right)\) at rest. (Assume that the gold \(n u-\) cleus remains at rest during the collision.) (b) What minimum initial speed must the alpha particle have in order to approach to a distance of \(300 \mathrm{fm}\) ?

In addition to the radioactive nuclei included in the natural decay series, there are several other radioactive nuclei that occur naturally. One is \({ }^{147} \mathrm{Sm}\), which is \(15 \%\) naturally abundant and has a half-life of approximately \(1.3 \times 10^{10}\) years. Calculate the number of decays per second per gram (due to this isotope) in a sample of natural samarium. The atomic weight of samarium is \(150.4\). (Activity per unit mass is called specific activity.)

In a piece of rock from the Moon, the \({ }^{87} \mathrm{Rb}\) content is assessed to be \(1.82 \times 10^{10}\) atoms per gram of material and the \({ }^{87} \mathrm{Sr}\) content is found to be \(1.07 \times 10^{9}\) atoms per gram. (a) Determine the age of the rock. (b) Could the material in the rock actually be much older? What assumption is implicit in the use of the radioactive dating method? (The relevant decay is \({ }^{87} \mathrm{Rb} \rightarrow{ }^{87} \mathrm{Sr}+\mathrm{e}^{-}\). The half-life of the decay is \(4.8 \times 10^{10}\) yr.)

A radioactive nucleus with decay constant \(\lambda\) decays to a stable daughter nucleus. (a) Show that the number of daughter nuclei, \(N_{2}\), increases with time according to the expression $$ N_{2}=N_{01}\left(1-e^{-\lambda t}\right) $$ where \(N_{01}\) is the initial number of parent nuclei. (b) Starting with \(10^{6}\) parent nuclei at \(t=0\), with a halflife of \(10 \mathrm{~h}\), plot the number of parent nuclei and the number of daughter nuclei as functions of time over the interval 0 to \(30 \mathrm{~h}\).

As part of his discovery of the neutron in 1932 , James Chadwick determined the mass of the newly identified particle by firing a beam of fast neutrons, all having the same speed, at two different targets and measuring the maximum recoil speeds of the target nuclei. The maximum speeds arise when an elastic head-on collision occurs between a neutron and a stationary target nucleus. (a) Represent the masses and final speeds of the two target nuclei as \(m_{1}, v_{1}, m_{2}\), and \(v_{2}\) and assume Newtonian mechanics applies. Show that the neutron mass can be calculated from the equation $$ m_{n}=\frac{m_{1} v_{1}-m_{2} v_{2}}{v_{2}-v_{1}} $$ (b) Chadwick directed a beam of neutrons (produced from a nuclear reaction) on paraffin, which contains hydrogen. The maximum speed of the protons ejected was found to be \(3.3 \times 10^{7} \mathrm{~m} / \mathrm{s}\). Since the velocity of the neutrons could not be determined directly, a second experiment was performed using neutrons from the same source and nitrogen nuclei as the target. The maximum recoil speed of the nitrogen nuclei was found to be \(4.7 \times 10^{6} \mathrm{~m} / \mathrm{s}\). The masses of a proton and a nitrogen nucleus were taken as \(1 \mathrm{u}\) and \(14 \mathrm{u}\), respectively. What was Chadwick's value for the neutron mass?

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