Question

Radioactive cobalt-60 is used to study defects in vitamin \(\mathbf{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \(^{60} \mathrm{Co}=\) \(\left.59.9338 \mathrm{u} ;^{1} \mathrm{H}=1.00782 \mathrm{u}\right) ?\) What is the de Broglie wave-length of the emitted particle if it has a velocity equal to \(0.90 c\) where \(c\) is the speed of light?

Short answer

The emitted particle during the nuclear synthesis of cobalt-60 is an electron (\(\beta^-\)). The binding energy per nucleon for cobalt-60 is approximately \(1.38\times 10^{-12}\, \textrm{J/nucleon}\). The de Broglie wavelength of the emitted particle with a velocity of \(0.90c\) is approximately \(2.699\times 10^{-12}\, \textrm{m}\).

Step-by-step solution

Analyze the nuclear equation

By analyzing the overall nuclear reaction, we'll be able to determine which particle is emitted during the synthesis. The overall reaction is given as: \(^{58}\textrm{Fe + 2}\,^{1}\textrm{n -> }^{60}\textrm{Co + X}\) where X is the unknown emitted particle. We can determine the particle by balancing the nucleon numbers (mass numbers) and proton numbers (atomic numbers). We have: Mass numbers: 58 + 2 = 60 + mass number of X Atomic numbers: 26 + 0 = 27 + atomic number of X Solving for the unknowns: mass number of X = 0 atomic number of X = -1 This implies the particle emitted is an electron, usually represented as \(\beta^-\) or \(^{0}_{-1}\textrm{e}\).

Calculate the binding energy per nucleon for cobalt-60

The binding energy per nucleon can be calculated using the formula: \(BE = \frac{(\Delta m)c^2}{A}\) where \(\Delta m\) is the mass defect, \(c\) is the speed of light, and \(A\) is the mass number. First, let's find the mass defect (\(\Delta m\)): \(\Delta m = (Mass_{\textrm{Fe}} + 2 \times Mass_{\textrm{n}}) - Mass_{\textrm{Co}}\) Using the provided atomic masses and substituting the values: \(\Delta m = \left(\frac{58}{26}\times 1.00782\textrm{u} + 2 \times 1.00782\textrm{u}\right) - 59.9338\textrm{u}\) \(\Delta m \approx 0.09202\,\textrm{u}\) Now we can calculate the binding energy: \(BE = \frac{(0.09202\,\textrm{u})\times (9\times 10^{13}\,\textrm{J})}{60}\) \(BE \approx 1.38\times 10^{-12}\,\textrm{J/nucleon}\)

Compute the de Broglie wavelength of the emitted particle

To calculate the de Broglie wavelength of the emitted particle (electron), we'll use the formula: \(\lambda = \frac{h}{p}\) where \(h\) is the Planck's constant, and \(p\) is the momentum of the particle, which we calculate using its mass and velocity. Since the mass of an electron is \(m_{e}=9.109\times 10^{-31}\textrm{kg}\) and its velocity is given as \(v=0.90c\), the momentum can be calculated as: \(p = m_{e}v\) \(p = (9.109\times 10^{-31}\textrm{kg})(0.90 \times 3\times 10^8\,\textrm{m/s})\) \(p \approx 2.454\times 10^{-22}\,\textrm{kg m/s}\) Now, we can calculate the de Broglie wavelength using Planck's constant \(h = 6.626\times 10^{-34}\,\textrm{Js}\): \(\lambda = \frac{6.626\times 10^{-34}\,\textrm{Js}}{2.454\times 10^{-22}\,\textrm{kg m/s}}\) \(\lambda \approx 2.699\times 10^{-12}\,\textrm{m}\) #Answer# 1. The emitted particle is an electron (\(\beta^-\)). 2. The binding energy per nucleon for cobalt-60 is approximately \(1.38\times 10^{-12}\,\textrm{J/nucleon}\). 3. The de Broglie wavelength of the emitted particle is approximately \(2.699\times 10^{-12}\,\textrm{m}\).

Key concepts

Radioactive Cobalt-60

Radioactive isotopes such as cobalt-60 play a significant role in both scientific research and practical applications. Cobalt-60, a synthetic isotope, is commonly used in medicine for radiation therapy, and in industry for radiography and sterilization. Its synthesis from iron-58 involves neutron capture, a common nuclear reaction in both stars and reactors.

When neutrons are absorbed by a nucleus, they can initiate a series of changes that lead to the emission of particles. In the case of cobalt-60 synthesis from iron-58, the iron nucleus captures two neutrons and transforms into a cobalt-60 nucleus. During this process, an additional particle, beta radiation in the form of an electron, is emitted. This particle emission is crucial because it helps to maintain the conservation of energy and particle number in nuclear reactions.

Nuclear synthesis of isotopes like cobalt-60 allows scientists to tailor make isotopes for specific purposes. The understanding of these processes is essential for fields ranging from nuclear physics to medical diagnostics and treatment.

Binding Energy per Nucleon

Binding energy per nucleon is a fundamental concept in nuclear physics, providing insight into the stability of a nucleus. Put simply, it is the energy that would be required to disassemble a nucleus into its individual protons and neutrons. The greater this energy, the more stable the nucleus is considered to be.

When considering isotopes like cobalt-60, the binding energy per nucleon tells us how tightly bound the nucleons are within the nucleus. The calculation of binding energy generally involves understanding the mass defect, which is based on the principle that mass and energy are equivalent as per Einstein's famous equation, E=mc².

Understanding binding energy per nucleon helps to explain why certain isotopes undergo radioactive decay, releasing energy, while others remain stable. It provides critical information for applications that range from understanding stellar processes to the design of nuclear reactors and development of new medical treatments.

De Broglie Wavelength

The de Broglie wavelength is a foundational concept in quantum mechanics, which describes the wave-like behavior of particles. According to de Broglie, all particles exhibit wave properties, and their wavelengths can be calculated if the momentum of the particle is known, using the formula \[\begin{equation}\lambda = \frac{h}{p},\end{equation}\]where \(h\) is the Planck constant and \(p\) is the particle's momentum.

In the nuclear synthesis of cobalt-60, when calculating the de Broglie wavelength of an emitted electron that moves with a significant fraction of the speed of light (0.90 c), this wave-particle duality becomes particularly relevant. The result allows physicists to predict the behavior of the electron, guiding the design of experiments and new technologies that rely on high-velocity subatomic particles.