Question

Radioactive cobalt-60 is used to study defects in vitamin \(\mathrm{B}_{12}\) 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: \({ }^{60} \mathrm{Co}=\) 59.9338 u; \({ }^{1} \mathrm{H}=1.00782 \mathrm{u}\) )? What is the de Broglie wavelength 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 in the nuclear synthesis of cobalt-60 is a beta particle (electron or positron). The binding energy per nucleon for cobalt-60 is approximately \(2.2 \times 10^{-13}\ \mathrm{J/nucleon}\), and the de Broglie wavelength of the emitted particle is approximately \(2.42 \times 10^{-15}\ \mathrm{m}\).

Step-by-step solution

Identify the emitted particle in the nuclear synthesis of cobalt-60

The overall nuclear synthesis reaction can be represented as: $$\,^{58}\mathrm{Fe} + 2\: ^1\mathrm{n} \rightarrow \,^{60}\mathrm{Co} + X$$ To find the emitted particle X, we can use conservation of nucleons, which states that the total number of protons and neutrons is conserved in a nuclear reaction. Therefore: $$58 + 2 \mathrm{n} \rightarrow 60 + X$$ We can see that two nucleons are converted into a single particle, which means that the emitted particle X is a beta particle (electron or positron).

Calculation of binding energy per nucleon

The binding energy per nucleon can be calculated using the mass difference between the interacting particles and the resulting particle and the conversion factor from atomic mass units to energy, which is known as the atomic mass constant \(E = mc^2\). The formula for binding energy is: $$E_B = \dfrac{(m_{initial} - m_{final})c^2}{A_{final}}$$ Here, \(m_{initial}\) represents the initial mass (mass of iron-58 + 2 * mass of neutron), \(m_{final}\) represents the final mass (mass of cobalt-60), and \(A_{final}\) is the final number of nucleons (60). We can use the given atomic masses to calculate the binding energy per nucleon for cobalt-60. The given masses are: - Mass of cobalt-60: 59.9338 u - Mass of neutron: 1.00782 u First we calculate, the mass of the initial nucleus: $$m_{initial} = 58 + 2 * 1.00782 = 60.01564\ \mathrm{u}$$ Now, we can calculate the binding energy: $$E_B = \dfrac{(60.01564 - 59.9338) * \big( \dfrac{931.49 \ \mathrm{MeV}}{1\ \mathrm{u}} \big)}{60}$$ $$E_B \approx 1.37\ \mathrm{MeV/nucleon}$$ Converting MeV to Joules: $$E_B = 1.37 * 1.60219 \times 10^{-13}\ \mathrm{J/nucleon}$$ $$E_B \approx 2.2 \times 10^{-13}\ \mathrm{J/nucleon}$$

Calculation of de Broglie wavelength of the emitted particle

To calculate the de Broglie wavelength of the emitted beta particle, we can use the de Broglie wavelength formula: $$\lambda = \dfrac{h}{p}$$ where h is the Planck's constant and p is the momentum of the particle. The momentum of the particle can be calculated using the formula: $$p = \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}$$ where m is the mass of the particle, and v is its velocity. The mass and charge of the emitted beta particle can be either an electron (negative) or positron (positive). However, since these particles have the same mass, it won't affect the calculation of wavelength. As given, the velocity of the emitted particle is 0.90 times the speed of light so, $$v = 0.9c$$ the mass of the electron (or positron) is approximately \(9.109 \times 10^{-31}\ \mathrm{kg}\). Now, we can calculate the momentum of the emitted beta particle: $$p = \dfrac{9.109 \times 10^{-31} * (0.9c)}{\sqrt{1-(0.9)^2}}$$ Now we can use this to find the de Broglie wavelength: $$\lambda = \dfrac{6.626 \times 10^{-34}\ \mathrm{J*s}}{p}$$ After plugging in the values, the calculated de Broglie wavelength of the emitted beta particle is approximately to: $$\lambda \approx 2.42 \times 10^{-15}\ \mathrm{m}$$ To summarize, the emitted particle is a beta particle (electron or positron), the binding energy per nucleon for cobalt-60 is approximately \(2.2 \times 10^{-13}\ \mathrm{J/nucleon}\), and the de Broglie wavelength of the emitted particle is approximately \(2.42 \times 10^{-15}\ \mathrm{m}\).

Key concepts

Nuclear Reactions

Nuclear reactions play a critical role in chemistry and physics, involving changes in an atom's nucleus that can lead to the transformation of elements. In educational settings, students explore these processes through examples such as the creation of cobalt-60. This isotope is synthesized through a nuclear reaction where iron-58 absorbs neutrons and undergoes a transformation.

Understanding the balance of the nucleons—protons and neutrons—is essential to grasp these reactions. For instance, conserving the total number of nucleons is a crucial principle, assisting students to deduce the identity of particles emitted during reactions. In our cobalt-60 example, two neutrons are absorbed by iron-58, and a beta particle is emitted. This conservation law enables learners to predict outcomes of complex nuclear processes.

Binding Energy per Nucleon

The concept of binding energy per nucleon is fundamental to grasp nuclear stability and why certain nuclei are more stable than others. This idea tells us how much energy is required to separate a nucleon from the nucleus.

In educational exercises, we often calculate the binding energy per nucleon to evaluate nuclear stability. The calculations involve determining the mass defect—the difference in mass between the bound nucleus and the sum of its separate nucleons. Students learn to use this mass defect along with Einstein's mass-energy equivalence, represented by the iconic equation, E = mc^2, to calculate the binding energy per nucleon. The larger the binding energy per nucleon, the more stable the nucleus. For instance, calculating the binding energy for cobalt-60 allows students to determine its stability compared to other isotopes.

de Broglie Wavelength

The de Broglie wavelength is a concept that introduces students to the wave-like properties of particles, which is a cornerstone of quantum mechanics. According to de Broglie's hypothesis, every moving particle has an associated wave with a wavelength given by λ = h/p, where h is Planck's constant and p is the particle's momentum.

In practice, this means that even particles with mass, like electrons, can exhibit wave-like behaviors. This concept challenges classical mechanics and is fascinating for learners as they delve into topics such as electron diffraction and the wave-particle duality. When students calculate the de Broglie wavelength for a beta particle emitted during the formation of cobalt-60, they get a glimpse into the quantum world where particles move at significant fractions of the speed of light.

Beta Particle Emission

Beta particle emission is a type of radioactive decay where an unstable nucleus transforms into a more stable one by emitting a beta particle. A beta particle can be an electron or its antiparticle, a positron. This process involves the transformation of a neutron into a proton (or vice versa), which leads to a change in the element's atomic number.

Understanding beta decay is important for students to comprehend how certain isotopes, such as cobalt-60, can form and then subsequently decay. In classroom problems, exploring beta particle emission helps illustrate the conservation of charge and nucleons, introduces the idea of neutrinos (which are often emitted alongside beta particles), and ties in with lessons on nuclear synthesis and stability. Students discover the interconnectedness of these nuclear phenomena through exercises like the synthesis of cobalt-60.