One of the decays of potassium (A=40, Z=19) results in an excited argon atom with excess energy of 1.50 Mev. In order to be stable, it emits a gamma photon. What frequency and wavelength has this gamma photon?

The nucleus of the atoms are usually in excited states after performing a beta decay. By emitting a photon, the nucleus recover stability. The emitted photon has the same energy as the excess energy of the excited nucleus. To calculate the frequency we need to use Planck's relation E=hf (where h is Planck's constant). To calculate the wavelength we realize the photon can behave like a wave travelling at speed of light, so we can use c=fλ. The results are found by f=E/h=3.63x10^20 and λ=c/f=8.27x10^(-13) . Note that we need to use SI units so we need to transform the energy from Mev to J.