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) $^{58}\textrm{Fe}$