We will write an equation for** sodium 22** undergoing **electron capture**

Recall that in an electron capture or *electron absorption reaction* our electron particle is a reactant and not a product.

It is the opposite of beta decay. The equation will look like this:

$\overline{){}_{\mathbf{atomic}\mathbf{}\mathbf{\#}}{}^{\mathbf{\#}\mathbf{}\mathbf{neutrons}}{\mathbf{E}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{}_{\mathbf{-}\mathbf{1}}{}^{{\mathbf{0}}}{\mathbf{e}}{\mathbf{\to}}{}_{{\mathbf{atomic}}{\mathbf{}}{\mathbf{\#}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{1}}}{}^{\mathbf{\#}\mathbf{}\mathbf{neutrons}}{\mathbf{X}}\mathbf{}}$* *

***Symbols for E and X do not represent actual elements *

* *****Atomic # = # of protons + # of electrons (neutral element)

***# Neutrons can be found based on the name of Isotope *

To write the equation for **Sodium 22** undergoing electron capture, we need to find the # of electrons from its **Atomic #** based on the Periodic Table of elements.

What is the equation for sodium 22 undergoing electron capture?

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