Final answer:
Iodine-131 undergoes beta decay to become xenon, with the emission of a beta particle (an electron) and sometimes a gamma ray. The correct answer is d. beta decay.
Step-by-step explanation:
Iodine-131 (atomic number 53) undergoes beta decay when it transforms into xenon (atomic number 54). In this process, iodine-131 emits a beta particle (an electron) and sometimes a gamma ray is emitted simultaneously. The balanced nuclear equation representing this decay is:
131I53 → 131Xe54 + 0−e1 (beta particle).
This is indicative of beta decay, as there is an increase in the atomic number by one, which occurs when a neutron in the iodine nucleus is converted into a proton, and an electron is emitted. The emitted beta particle is the electron noted in the equation.
Iodine-131 commonly undergoes beta decay. In beta decay, a beta particle (an electron or a positron) is emitted from the nucleus. The balanced equation for the decay of iodine-131 is:
I-131 → Xe-131 + e-
The daughter isotope formed is xenon-131, and a gamma ray may also be emitted simultaneously with the beta particle.