Respuesta :
Answer : The correct option is, (C) [tex]^{58}\textrm{Fe}[/tex]
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,
[tex]_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma[/tex]
(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,
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e[/tex]
(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.
[tex]_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma[/tex]
Process 2 : Beta emission.
[tex]_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e[/tex]
Process 3 : Neutron capture.
[tex]_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma[/tex]
From this we conclude that, the initial reactant in the production of cobalt-60 is [tex]_{26}^{58}\textrm{Fe}[/tex]
Hence, the correct option is, (C) [tex]^{58}\textrm{Fe}[/tex]
