Final answer:
The energy of a photon from cobalt-60 gamma radiation with a wavelength of 1.00 × 10^-9 meters is 1.988 x 10^-16 joules.
Step-by-step explanation:
The question involves calculating the energy of a photon from the gamma radiation emitted by cobalt-60, which has a given wavelength. The energy (E) of a photon can be calculated using the equation E = hc/λ, where h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light in a vacuum (3.00 x 10^8 m/s), and λ is the wavelength of the photon. In this case, the wavelength is given as 1.00 × 10^-9 meters (nm).
By substituting the given values into the equation, we calculate the energy:
E = (6.626 x 10^-34 J·s) x (3.00 x 10^8 m/s) / (1.00 x 10^-9 m)
E = (6.626 x 10^-34 x 3.00 x 10^8) / 10^-9 J
E = 1.988 x 10^-16 joules per photon.
This is the energy of a single photon of gamma radiation emitted by a cobalt-60 source.