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

The fraction of electrons removed from the copper ball can be calculated by comparing its net charge to the charge of a single electron. Using the provided information, we can find that approximately 2.5 x 10^10 electrons have been removed from the ball. To determine the fraction, we need to compare this number to the total number of electrons in the ball, which can be calculated using the mass, density, and atomic mass of copper. By plugging in the values, we can find the fraction of electrons removed.

Step-by-step explanation:

To determine the fraction of electrons that have been removed from the copper ball, we need to compare the net charge of the ball to the charge of a single electron. The net charge of the ball is 40 nC, which is equivalent to 40 x 10^-9 C. The charge of a single electron is 1.60 x 10^-19 C.

We can calculate the number of electrons that have been removed using the formula:

Number of electrons removed = Net charge of the ball / Charge of a single electron

Number of electrons removed = (40 x 10^-9 C) / (1.60 x 10^-19 C) = 2.5 x 10^10 electrons

To find the fraction of electrons removed, we need to compare the number of electrons removed to the total number of electrons in the ball. The total number of electrons in the ball can be calculated using the formula:

Total number of electrons = Number of copper atoms x Number of electrons per copper atom

The number of copper atoms can be calculated using the formula:

Number of copper atoms = Mass of the ball / Atomic mass of copper

The mass of the ball can be calculated using the formula:

Mass of the ball = Volume of the ball x Density of copper

Given that the diameter of the ball is 2.0 mm, the volume of the ball can be calculated using the formula for the volume of a sphere:

Volume of the ball = (4/3) x pi x (radius)^3

As the ball is a sphere, the radius is half the diameter, so the radius is 1.0 mm or 1 x 10^-3 m.

Using the given density of copper (8900 kg/m^3) and atomic mass of copper (63.5 g/mol), we can now calculate the fraction of electrons removed:

Fraction of electrons removed = Number of electrons removed / Total number of electrons = (2.5 x 10^10 electrons) / (Number of copper atoms x Number of electrons per copper atom)

Answer 2

0.02442 × 10⁻⁹

Step-by-step 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 =
`(40*10^(-9))/(1.6*10^(-19))`

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =
`(4)/(3)\pi((d)/(2))^3` × 8900

or

Mass of copper ball =
`(4)/(3)\pi((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 =
`( 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 =
`(25*10^(10))/(1023.37*10^(19))`

or

The fraction of electrons removed = 0.02442 × 10⁻⁹