Question

(a) How many electrons would have to be removed from a penny to leave it with a charge of +1.0×10−7 C? [4]
(b) To what fraction of the electrons in the penny does this correspond? [6]
[A penny has a mass of 3.11 g; assume it is made entirely of copper.]

Answer 1

Answer:

The answer is 6.25 × 10 ^11  electrons  and 7.31 × 10 ^− 13

Explanation:

The charge on an electron is  

1.6 × 10 ^− 19 C .

So for part (a) we can determine that the number of electrons would be:

1.0 × 10 − ^7 /  1.6 × 10^ − 19  =  6.25 × 10 ^11  electrons

For part b, we need to know how may electrons are in the penny. The relative atomic mass of copper is 63.546. By definition, 63.546g of copper would contain a number of atoms equal to Avogadro's constant which is  

6.02 × 10^ 23  m o l − 1 .  

So in 3.11 g there would be the following number of atoms:  

3.11 / 63.546 × ( 6.02 × 10 ^23 )  =  2.95 ×10^ 22  atoms

The atomic number of copper is 29, so a neutral copper atom has 29 electrons.  

Hence the fraction of electrons that would need to be removed would be:  

6.25 × 10 ^11  / 29 × ( 2.95 × 10 ^22  ) =  7.31 × 10 ^− 13