Question

Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial reactant in the production of cobalt-60 is ________.a. 59Co b. 56Fe c. 58Fe d. 61Co e. 60Fe

Answer 1

Answer : The correct option is, (C)
`^(58)\textrm{Fe}`

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

`_Z^A\textrm{X}+_(0)^1\textrm{n}\rightarrow _(Z)^(A+1)\textrm{X}+\gamma`

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

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

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

`_(26)^(58)\textrm{Fe}+_(0)^1\textrm{n}\rightarrow _(26)^(59)\textrm{Fe}+\gamma`

Process 2 : Beta emission.

`_(26)^(59)\textrm{Fe}\rightarrow _(27)^(59)\textrm{Co}+_(-1)^0e`

Process 3 : Neutron capture.

`_(27)^(59)\textrm{Co}+_(0)^1\textrm{n}\rightarrow _(26)^(60)\textrm{Co}+\gamma`

From this we conclude that, the initial reactant in the production of cobalt-60 is
`_(26)^(58)\textrm{Fe}`

Hence, the correct option is, (C)
`^(58)\textrm{Fe}`