Respuesta :
The daughter nuclide formed when copper undergo positron emission is
nickel, with atomic number z= 28.
let us show the work
63 29Cu →+10 e + 63 28
In this decay , the a proton in the nucleus of copper changes to a neutron, a
positron is emitted, the atomic number reduces by one why the atomic mass
remain unchanged.
Answer: [tex]_{28}^{59}\textrm {Ni}[/tex]
Explanation:-
Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as [tex]\beta ^+[/tex]-decay. In this the mass number remains same.
General representation of an element is given as:
where,
Z represents Atomic number
A represents Mass number
X represents the symbol of an element
The chemical equation for positron emission is represented as:
[tex]_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e[/tex]
[tex]_{29}^{59}\textrm{Cu}\rightarrow _{28}^{59}\textrm{Ni}+_{+1}^0e[/tex]