# Finding the fraction of electrons lost

• jheld

## Homework Statement

A small piece of aluminum(atomic number 13) contains 1015 . (The atomic number is the number of protons; it determines the (positive) electric charge of the nucleus and, thus, the number of electrons in a neutral atom.) If the piece of aluminum has a net positive charge of 3.0 micro Coulombs, what fraction of the electrons that the aluminum had when it was neutral would have had to be lost?

## Homework Equations

I am not sure exactly what I need as far as equations go.
I know what the answer is, but I am unsure of how to get it.

## The Attempt at a Solution

I don't know how to work this problem fully. I have an idea, but I get stuck along the way.

A small piece of aluminum(atomic number 13) contains 1015 . (The atomic number is the number of protons; it determines the (positive) electric charge of the nucleus and, thus, the number of electrons in a neutral atom.) If the piece of aluminum has a net positive charge of 3.0 micro Coulombs, what fraction of the electrons that the aluminum had when it was neutral would have had to be lost?.

Hi jheld!

When it's neutral, it has 1015 atoms, each of which has 13 electrons …

so how much positive and negative charge (equal, of course) does it have?

And so if it now has a net positive charge of 3.0 µC, what fraction of the electrons has been lost?

Good point!
So I can say that with the positive charge increase...
(3*10^-6)/(13*1.6*10^-19*10^15) = some crappy decimal which comes out to be 3/2080.

thanks for the help :)