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 iron58 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: \({ }^{60} \mathrm{Co}=59.9338\) amu; \({ }^{1} \mathrm{H}=1.00782\) amu \()\) ? 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?

We have the overall reaction: iron58 + 2 neutrons → cobalt-60 + unknown_particle To identify the missing particle, we need to balance the equation. Iron-58 has 26 protons and 32 neutrons (58 = 26 + 32). Cobalt-60 has 27 protons and 33 neutrons (60 = 27 + 33). Hence, the nuclear reaction would be: $$ \text{iron-58}+2\text{ neutrons}=\text{cobalt-60}+\text{unknown_particle} $$ Now, we need to balance the number of protons and neutrons on each side: Protons: 26 (iron-58) = 27 (cobalt-60) + x (unknown_particle) Neutrons: 32 (iron-58) + 2 = 33 (cobalt-60) + y (unknown_particle) Solving these equations, we get x = 1 and y = 1. Therefore, the unknown particle has one proton and one neutron, which is deuterium (heavy hydrogen). So the particle emitted in the nuclear synthesis is deuterium.

Now, we determine the binding energy per nucleon for cobalt-60. First, we need to find the mass defect. Mass defect = Total mass of the nucleus - Mass of cobalt-60 The total mass of the nucleus is the sum of the masses of 27 protons and 33 neutrons. Mass of a proton = Mass of hydrogen-1 = 1.00782 amu Mass of a neutron ≈ Mass of a proton ≈ 1.00782 amu Total mass of the nucleus ≈ (27 × 1.00782) + (33 × 1.00782) Mass defect ≈ ((27 × 1.00782) + (33 × 1.00782)) - 59.9338 Next, we convert the mass defect to energy using the mass-energy equivalence formula: $$ E = mc^2 $$ Where E is the energy, m is the mass defect, and c is the speed of light. First, let's convert the mass defect to kg: Mass defect (kg) = mass defect (amu) × 1.6605 × 10^{-27} Now, we can find the binding energy: Binding energy = mass defect (kg) × c^2 $$ E = (mass\:defect)\times(1.6605\times10^{-27}\:kg)\times(3\times10^{8}\:m/s)^2 $$ Finally, we find the binding energy per nucleon: Binding energy per nucleon = Binding energy / 60

We have the de Broglie wavelength formula: $$ \lambda = \frac{h}{mv} $$ Where λ is the de Broglie wavelength, h is Planck's constant (6.626 × 10^{-34} Js), m is the mass of the particle, and v is its velocity. The mass m in this case is the mass of the emitted deuterium and its velocity is 0.9c. Mass of deuterium = Mass of 1 proton + Mass of 1 neutron ≈ 2 × Mass of a proton Convert the mass of deuterium to kg: Mass of deuterium (kg) = 2 × mass of a proton (amu) × 1.6605 × 10^{-27} Velocity of deuterium = 0.9c = 0.9 × 3 × 10^8 m/s Now, we can calculate the de Broglie wavelength: $$ \lambda = \frac{6.626\times10^{-34}\:Js}{(2\times1.6605\times10^{-27})\times(0.9\times3\times10^8\:m/s)} $$ From these steps, we can now find the emitted particle in the nuclear synthesis, binding energy per nucleon for cobalt-60, and the de Broglie wavelength of the emitted particle.