Final answer:
The frequency of a photon emitted by a radium-266 atom is equal to the energy difference between its energy levels. This is because the photon's energy, which determines its frequency, is directly proportional to this energy difference and is not related to the atom's atomic or mass number, or the speed of light.
Step-by-step explanation:
The main answer to the question 'An radium-266 atom may emit a photon of light with a frequency' is c) Equal to the energy difference between its energy levels. When an atom transitions between different energy levels, it emits or absorbs photons with a frequency that corresponds to the energy difference between those levels. This follows from the principle that the energy of a photon (E) is related to its frequency (ν) by the equation E=hν, where h is Planck's constant. The frequency of the photon is not directly related to properties such as the atom's atomic number, mass number, or the speed of light, which eliminates options a, b, and d, respectively.
For instance, electromagnetic radiation, which includes photons, all travel at the speed of light (c) in a vacuum, but they differ in wavelength and frequency. The relationship between the speed of light, wavelength (λ), and frequency is given by the equation c = λν. This equation shows that the frequency is inversely proportional to the wavelength.