Question

What is the daughter nucleus produced when 59cu undergoes positron emission?

Answer 1

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 2

Answer:
`_(28)^(59)\textrm {Ni}`

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
`\beta ^+`-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:

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

`_(29)^(59)\textrm{Cu}\rightarrow _(28)^(59)\textrm{Ni}+_(+1)^0e`