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: \(^{60} \mathrm{Co}=\) $59.9338 \mathrm{u} ;^{-1} \mathrm{H}=1.00782$ 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?

First, we need to examine the following reaction for the nuclear synthesis involving iron-58 and two neutrons: \[^{58} \mathrm{Fe} + 2^{1} \mathrm{n} \rightarrow ^{60} \mathrm{Co} + X\] where X is the unknown emitted particle. Let's find the missing particle by calculating the mass number and atomic number of the missing particle, X: Mass number of X = Mass number of reactants - Mass number of products Atomic number of X = Atomic number of reactants - Atomic number of products Reactants atomic number: \(Z_{Fe} = 26\), \(Z_n = 0\) Reactants mass number: \(A_{Fe} = 58\), \(A_n = 1\) Product atomic number: \(Z_{Co} = 27\) Product mass number: \(A_{Co} = 60\) Now, let's substitute the values in the equations and find the values for X: Mass number of X = (58 + 2 - 60) = 0 Atomic number of X = (26 + 0 - 27) = -1 The emitted particle, X, has a mass number of 0 and an atomic number of -1, which corresponds to an electron or a beta particle, (\(\beta^{-}\)). Hence, the nuclear synthesis equation is: \[^{58} \mathrm{Fe} + 2^{1} \mathrm{n} \rightarrow ^{60} \mathrm{Co} + \beta^{-}\]

As we know the atomic masses of cobalt-60 and hydrogen atom, we can calculate the binding energy per nucleon using the formula: Binding energy = \((Zm_H + Nm_n - m)(c^2)\) where \(Z\) is the atomic number of cobalt-60, \(N\) is the number of neutrons, \(m_H\) is the mass of 1 hydrogen atom, \(m_n\) is the mass of 1 neutron, and \(m\) is the given atomic mass of cobalt-60. Now, let's substitute the values: \(Z = 27\) \(N = 33\) \(m_H = 1.00782\ \mathrm{u}\) \(m_n = 1.00867\ \mathrm{u}\) \(m = 59.9338\ \mathrm{u}\) Binding energy = \(((27)(1.00782) + (33)(1.00867) - 59.9338)(c^2)\) Converting atomic mass units (u) to kilograms (kg): Binding energy = \(((27)(1.00782 \times 1.66 \times 10^{-27}) + (33)(1.00867 \times 1.66 \times 10^{-27} ) - (59.9338 \times 1.66 \times 10^{-27}))(9 \times 10^{16})\). Binding energy = \(1.60287 \times 10^{-13}\ \mathrm{J}\). To find the binding energy per nucleon, we divide the total binding energy by the number of nucleons: Binding energy per nucleon = \(\frac{1.60287 \times 10^{-13}}{60}\ \mathrm{J} / \mathrm{nucleon}\).

The de Broglie wavelength is given by: \(\lambda = \frac{h}{mv}\), where \(h\) is the Planck's constant, \(m\) is the mass of the particle, and \(v\) is the velocity of the particle. We know the emitted particle is a beta particle (electron), and its mass (\(m_e\)) is approximately \(9.11 \times 10^{-31}\ \mathrm{kg}\), and the velocity is \(0.90c\). The speed of light \(c = 3 \times 10^8\ \mathrm{m/s}\). So, the velocity of electron is \(v_e = (0.90) (3 \times 10^8) = 2.7 \times 10^8\ \mathrm{m/s}\). Planck's constant (\(h\)) is \(6.626 \times 10^{-34}\ \mathrm{J\cdot s}\). Now, let's substitute the values: \(\lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.7 \times 10^8)}\). After calculating, we find the de Broglie wavelength of the emitted particle: \(\lambda = 2.7 \times 10^{-15}\ \mathrm{m}\).