Question

The isotope cobalt-60 is widely used in cancer Treatments as it produces gamma rays as well as beta particles. Which of the following shows the most accurate equation for the beta and gamma radiation produced by Cobalt-60?

Answer 1

Cobalt-60 undergoes beta decay which can be represented by the equation ^60_27Co \rightarrow ^60_28Ni + \beta^- + \gamma. It has a half-life of 5.27 years which means its activity is halved every 5.27 years and must be replaced regularly in medical treatments.

Step-by-step explanation:

Cobalt-60 is a synthetic radioisotope widely used in cancer treatment. It is produced through the neutron activation of Co-59, which results in Co-60. This isotope undergoes beta decay, emitting beta particles and gamma rays, which can be focused on small areas to treat cancer. The equation representing the decay of Co-60 is:

^60_27Co \rightarrow ^60_28Ni + \beta^- + \gamma

Here, Co-60 decays into Ni-60, and during this process, a beta particle (\beta^-) and gamma radiation (\gamma) are emitted. Cobalt-60 has a half-life of approximately 5.27 years, which means its activity, and thus its intensity of radiation, is halved every 5.27 years. This decay process follows first-order kinetics. Because of its radioactive nature, it must be replaced regularly in medical applications to maintain its effectiveness in cancer therapy.