Question

A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’ electrons, removed and placed 1.00 m above it, would support the weight of this coin?

A certain five cent coin contains 5.00 g of nickel. What
fraction of the nickel atoms’ electrons, removed and placed
1.00 m above it, would support the weight of this coin? The
atomic mass of nickel is 58.7, and each nickel atom contains
28 electrons and 28 protons.

Answer 1

When we strip electrons off nickel atoms, these become positively charged. The force of attraction between electrons placed above at a distance of `1m` and nickel ions in the coin is required to support weight of the coin.

Let `-q` be charge of the stripped electrons.
From Law of Conservation of charge
Charge on the coin `=+q`

Setting up the weight balancing equation

`mg=k_e(∣q|^2)/r^2`,
where charges are separated by a distance `r`, `m` is mass of coin, `g=9.81ms^-2` and `k_e≈8.99×10^9N·m^2C^-2`

Inserting given values in SI units to find out `|q|`

`5.00/1000xx9.81=8.99×10^9(∣q|^2)/1^2`
`=>∣q|=sqrt((5.00xx9.81)/(1000xx8.99×10^9))`
`=>∣q|=2.34xx10^-6C`

Charge `|q|` in terms of electronic charge units (`1.602 × 10^-19C`)`=(2.34xx10^-6)/(1.602 × 10^-19)=1.458xx10^13`

We know that `58.7g` of nickel will contain Avogadro's number of atom`=6.022×10^23`
`:.` Number of nickel atoms in the coin`=6.022×10^23xx5.00/58.7`

`=5.1295xx10^22`

Total number of electrons in the coin`=5.1295xx10^22xx28`

`=1.436xx10^24`

Fraction of nickel atoms’ electrons stripped`=(1.458xx10^13)/(1.436xx10^24)`

`=1.02xx10^-11`, rounded to two decimal places.