Respuesta :

W0lf93

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]