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Chapter 45: Problem 22

What mass of \(\frac{60}{27}\) Co has an activity of \(1.0 \mathrm{Ci}\) ? The half-life of cobalt-60 is \(5.25\) years.

Short Answer

Expert verified
The mass of \( ^{60}_{27} \text{Co} \) with an activity of 1.0 Ci is approximately 8.73 g.

Step by step solution

01

Convert Activity to Decays per Second

First, we convert the given activity from curies (Ci) to decays per second (Bq). 1 Ci is equal to \(3.7 \times 10^{10}\) decays per second. Therefore:\[1.0 \text{ Ci} = 1.0 \times 3.7 \times 10^{10} \text{ Bq} = 3.7 \times 10^{10} \text{ decays per second}\]
02

Calculate the Decay Constant

The decay constant \( \lambda \) can be calculated using the half-life formula:\[\lambda = \frac{\ln(2)}{\text{Half-life}}\]Substitute the given half-life of cobalt-60:\[\lambda = \frac{\ln(2)}{5.25 \times 365 \times 24 \times 3600 \text{ seconds}}\]
03

Use Activity Formula to Find the Number of Atoms

The activity \( A \) is related to the number of atoms \( N \) by the formula:\[A = \lambda N\]Rearrange for \( N \):\[N = \frac{A}{\lambda}\]Substitute the values for \( A \) and \( \lambda \) calculated in the previous steps.
04

Convert Number of Atoms to Mass

Use Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \) and the molar mass of cobalt-60 (approximately \(59.9338 \text{ g/mol}\)) to convert the number of atoms to mass.First, find the number of moles \( n \):\[n = \frac{N}{N_A}\]Then calculate the mass \( m \):\[m = n \times 59.9338 \text{ g/mol}\]

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Key Concepts

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

Radioactive Decay
Radioactive decay is a fundamental concept in nuclear physics. It's the process by which an unstable atomic nucleus loses energy by emitting radiation. This emission can take the form of alpha particles, beta particles, or gamma rays.
The driving force behind radioactive decay is the pursuit of stability. An unstable nucleus wants to achieve a more stable configuration, which is why it releases energy.
Each radioactive isotope, or nuclide, decays at a unique and predictable rate, which is quantified by its half-life.
- **Spontaneous Process:** Radioactive decay occurs without any external influence. - **Release of Energy:** As decays happen, energy is released in the form of radiation. - **Transformation:** Through decay, a parent nuclide transforms into a daughter nuclide. Understanding radioactive decay helps in fields like medicine, archaeology, and energy production.
Cobalt-60
Cobalt-60 (\(^{60}_{27} \text{Co}\)) is a synthetic radioactive isotope of cobalt. It is commonly used in various applications due to its properties.
Cobalt-60 is notable for emitting gamma rays, which makes it valuable in medical and industrial settings.
- **Medical Applications:** Used in radiation therapy for cancer treatment.- **Industrial Use:** Employed for sterilization of medical equipment and food preservation.- **Scientific Research:** Utilized as a radiation source in laboratories.The isotope is produced by neutron activation of cobalt-59. As \(^{60}_{27} \text{Co}\) is unstable, it decays to nickel-60, releasing gamma radiation in the process. The half-life of cobalt-60 is significant because it determines how long it remains active and useful. The half-life of 5.25 years indicates that in this period, half of the \(^{60}_{27} \text{Co}\) nuclei would have decayed.
Half-life Calculation
Half-life is a crucial concept for understanding how long a radioactive substance remains active. It is the time required for half of the radioactive nuclei in a sample to decay.
This property is constant for any given isotope and is unaffected by environmental factors such as temperature or pressure.
To calculate the half-life, the decay constant \( \lambda \) is used. The relationship is given by the formula:
\[ \lambda = \frac{\ln(2)}{\text{Half-life}} \]- **Highlights of Half-Life:** - ***Predictability:** Helps estimate the duration an isotope can be usable. - ***Stability Insight:** Short half-lives indicate quick decay, while long half-lives imply slower decay. - ***Quantitative Measure:** Used to calculate remaining quantity of isotope over time.Understanding how to utilize half-life in calculations is essential for effective application in fields that involve radioactive materials, such as determining the dosage in medical treatments or analyzing radioactive isotopes in environmental studies.

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Most popular questions from this chapter

The half-life of carbon- 14 is \(5.7 \times 10^{3}\) years. What fraction of a sample of \({ }^{14} \mathrm{C}\) will remain unchanged after a period of five half-lives?

A nucleus \({ }_{n}^{M} \mathrm{P}\), the parent nucleus, decays to a daughter nucleus Dositron decay: $$ { }_{n}^{M} \mathrm{P} \rightarrow \mathrm{D}+{ }_{+1}^{0} e+{ }_{0}^{0} v $$ where \(v\) is a neutrino, a particle that has nearly zero mass and zero charge. ( \(a\) ) What are the subscript and superscript for D? (b) Prove that the mass loss in the reaction is \(M_{p}-M_{d}-2 m_{e}\), where \(M_{p}\) and \(M_{d}\) are the atomic masses of the parent and daughter. (a) To balance the subscripts and superscripts, we must have \({ }_{n-1} \mathrm{M} \mathrm{D}\). (b) The table of masses for the nuclei involved is \(\begin{array}{lc} & \text { Reactant mass } \\ M_{\mu} \mathrm{P} & M_{p}-n m_{e} \\ \text { TOTAL. } & M_{p}-n m_{e}\end{array}\) \({ }_{n-1}^{M} \mathrm{D}\) Product mass \(M_{d}-(n-1) m_{e}\) \(m_{e}\) \(\frac{\approx 0}{M_{d}-n m_{e}+2 m_{e}}\) ie \(0_{u}\) \(0_{0} v\) \(\begin{array}{rrr} & \pi\end{array}\) Subtraction gives the mass loss: $$ \left(M_{p}-n m_{e}\right)-\left(M_{d}-n m_{e}+2 m_{e}\right)=M_{p}-M_{d}-2 m_{e} $$ Notice how important it is to keep track of the electron masses in this and the previous problem.

In a mass spectrograph, the masses of ions are determined from their deflections in a magnetic field. Suppose that singly charged ions of chlorine are shot perpendicularly into a magnetic field \(B=0.15 \mathrm{~T}\) with a speed of \(5.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\). (The speed could be measured by use of a velocity selector.) Chlorine has two major isotopes, of masses \(34.97 \mathrm{u}\) and \(36.97 \mathrm{u}\). What would be the radii of the circular paths described by the two isotopes in the magnetic field? (See Fig. \(45-1 .\) ) The masses of the two isotopes are $$ m_{1}=(34.97 \mathrm{u})\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=5.81 \times 10^{-26} \mathrm{~kg} $$ $$ m_{2}=(36.97 \mathrm{u})\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=6.14 \times 10^{-26} \mathrm{~kg} $$ Because the magnetic force \(q v B\) must provide the centripetal force \(m v^{2} / r\), we have $$ r=\frac{m v}{q B}=\frac{m\left(5.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}{\left(1.6 \times 10^{-19} \mathrm{C}\right)(0.105 \mathrm{~T})}=m\left(2.98 \times 10^{24} \mathrm{~m} / \mathrm{kg}\right) $$ Substituting the values for \(m\) found above gives the radii as \(0.17 \mathrm{~m}\) and \(0.18 \mathrm{~m}\).

The radius of a carbon nucleus is about \(3 \times 10^{-15} \mathrm{~m}\) and its mass is \(12 \mathrm{u}\). Find the average density of the nuclear material. How many more times dense than water is this? $$ \begin{aligned} \rho &=\frac{m}{V}=\frac{m}{4 \pi r^{3} / 3}=\frac{(12 \mathrm{u})\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)}{4 \pi\left(3 \times 10^{-15} \mathrm{~m}\right)^{3} / 3}=1.8 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3} \\\ \frac{\rho}{\rho_{\text {water }}} &=\frac{1.8 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2 \times 10^{14} \end{aligned} $$

Neon- 23 beta-decays in the following way: $$ { }_{10}^{23} \mathrm{Ne} \rightarrow{ }_{11}^{23} \mathrm{Na}+{ }_{-1}^{0} e+{ }_{0}^{0} \bar{v} $$ where \(\bar{v}\) is an antineutrino, a particle with no charge and almost no mass. Depending on circumstances, the energy carried away by the antineutrino can range from zero to the maximum energy available from the reaction. Find the minimum and maximum KE that the beta particle \(_{-1}^{0} e\) can have. Pertinent atomic masses are \(22.9945 \mathrm{u}\) for \({ }^{23} \mathrm{Ne}\), and \(22.9898 \mathrm{u}\) for \({ }^{23} \mathrm{Na}\). The mass of the beta particle is \(0.00055 \mathrm{u}\). Note that the given reaction is a nuclear reaction, while the masses provided are those of neutral atoms. To calculate the mass lost in the reaction, subtract the mass of the atomic electrons from the atomic masses given. We have the following nuclear masses: which gives a mass loss of \(22.9945-22.9898=0.0047\) u. Since \(1.00 \mathrm{u}\) corresponds to \(931 \mathrm{MeV}\), this mass loss corresponds to an energy of \(4.4 \mathrm{MeV}\). The beta particle and antineutrino share this energy. Hence, the energy of the beta particle can range from zero to \(4.4 \mathrm{MeV}\).
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