When the nucleons rearrange in the following daughter nuclei, the energy changes by the amount shown and a \(\gamma\)-ray is emitted. Determine the frequency and wavelength of the \(\gamma\)-ray in each case: (a) cobalt-60, \(1.33 \mathrm{MeV}\); (b) arsenic-80, \(1.64 \mathrm{MeV}\); (c) iron-59, 1.10 MeV. (1 MeV \(=1.602 \times 10^{-13} \mathrm{~J}\).)

Nuclear decay is a process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This decay is a natural and spontaneous process that can lead to the production of a different element or a different isotope of the same element. In the context of gamma-ray emissions, which are a form of electromagnetic radiation, the decay often involves a transition of a nucleus from a higher to a lower energy state.

When this transition occurs, the difference in energy between the initial and final nuclear states is released in the form of a gamma-ray photon. This energy is what is referred to in the exercise when it mentions the change by the amount shown, for example, the 1.33 MeV for cobalt-60. The energy released during such nuclear transitions is usually quantified in units of electronvolts (eV) or its multiples, such as kiloelectronvolts (keV) or megaelectronvolts (MeV). Understanding nuclear decay is crucial as it sets the foundational knowledge needed for calculating the frequency and wavelength of the emitted gamma-rays from these events.

In the given problem, we see energy levels expressed in megaelectronvolts (MeV). However, to perform calculations involving Planck's law or to relate this energy to other forms of energy, we need to convert these values into joules (J), the standard unit of energy in the International System of Units (SI).

The conversion factor provided in the exercise, where 1 MeV equals 1.602 x 10^-13 J, is crucial for this transformation. Conversion of energy units is a common practice in physics because it allows for consistent and comparative analysis across different types of energy phenomena. The understanding of this conversion is a critical step for any student studying nuclear physics or working with energy at the quantum level because it bridges the gap between the microscopic world (where energies are often discussed in eV) and the macroscopic world (where joules are more commonly used).

Planck's law relates the energy of a photon to its frequency through the equation \(E=hf\), where \(E\) represents the energy of the photon in joules, \(h\) is Planck’s constant (\(6.626 \times 10^{-34} J\cdot s\)), and \(f\) is the frequency of the photon. By applying this equation, as seen in the solution steps for the exercise, we can calculate the frequency once the energy is known in joules.

To find the wavelength of the \(\gamma\)-ray, we use the speed of light equation \(c=\lambda f\) where \(c\) is the speed of light (approximately \(3 \times 10^{8} m/s\)), \(\lambda\) is the wavelength in meters, and \(f\) is the frequency in hertz (Hz). By isolating \(\lambda\), students can determine the wavelength with the previously found frequency. Understanding Planck's law and its application to gamma-ray frequency and wavelength calculation is essential in various fields of physics, including atomic, nuclear, and quantum physics, as well as in practical applications like medical imaging and astronomy.