Question

A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8900 kg/m^3.

Answer 1

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = $\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}$

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = $\frac{40\times10^{-9}}{1.6\times10^{-19}}$

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =  $\frac{4}{3}\pi(\frac{d}{2})^3$  × 8900

or

Mass of copper ball =  $\frac{4}{3}\pi(\frac{0.002}{2})^3$  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = $\frac{ 0.03726}{63.546}$

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = $\frac{25\times10^{10}}{1023.37\times10^{19}}$

or

The fraction of electrons removed = 0.02442 × 10⁻⁹