Question

60m27Co → 6027Co

Predict the type of radioactive emission produced from the decay of metastable cobalt-60 to cobalt-60. Describe this type of emission and its reaction to an electric field.

Answer 1

Answer: C (on usatestprep)

Step-by-step explanation:

Radioactive gamma decay is produced by the reaction. This neutral electromagnetic radiation allows the isotope to return to its ground state and is not attracted to the electric field.

Answer 2

It is gamma emission and there would be no reaction with the electric field.

Step-by-step explanation:

1.) Alpha decay: In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

`_Z^A\textrm{X}\rightarrow _(Z-2)^(A-4)+_2^4\alpha`

2.)Beta-decay: In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

`_Z^A\textrm{X}\rightarrow _(Z+1)^A\textrm{Y}+_(-1)^0\beta`

3.) Gamma ray emission: in this process, an unstable nuclei gives off excess energy by a spontaneous electromagnetic process and releases
`\gamma -radiations`. These radiations does not carry any charge and are electrically neutral.

`_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma`

4.) Positron decay: In this process, a proton gets converted to neutron and an electron neutrino and releases positron particles. This particle carries a charge of +1 units.

`_Z^A\textrm{X}\rightarrow _(Z-1)^A\textrm{Y}+_(+1)^0e`

The decay of metastable cobalt-60 to cobalt-60 is a gamma ray emission process. Since these radiations do not carry any charge, they do not react with the electric field.