Answer
(a) 60Co ⇾ 60Ni + e- + υe
(b)222Rn ⇾ 222Po + e- + υe
Work Step by Step
(a)The beta decay of Co- 60 can be represented by the following nuclear equation:
60Co ⇾ 60Ni + e- + υe
In this equation, 60Co represents the cobalt-60 isotope, which is undergoing beta decay. The product of the decay is the nickel-60 isotope (60Ni), an electron (e-), and an antineutrino (υe).
In the case of cobalt-60, the beta decay process transforms it into nickel-60 by decreasing the number of neutrons by one.
(b) The beta decay of radon-222 can be represented by the following nuclear equation:
222Rn ⇾ 222Po + e- + υe
In this equation, Rn-222 represents the radon-222 isotope, which is undergoing beta decay. The product of the decay is the polonium-222 isotope (222Po), an electron (e-), and an antineutrino (υe).
In the case of radon-222, the beta decay process transforms it into polonium-222 by decreasing the number of neutrons by two.
