Chapter 20, Problem 42A

To determine

The coulombs of charge on the electrons of a nickel.

Answer to Problem 42A

The coulombs of charge on the electrons of a nickel is $\text{2}\text{.24}\times {\text{10}}^5\text{ coulombs}$ .

Explanation of Solution

Given:

The mass of nickel is $5\text{ g}$ .

The mass of each mole of coin’s atoms is $62\text{ g}$ .

The number of electrons in each Cu atom is $29\text{ electrons}$ .

The number of electrons in each Ni atom is $28\text{ electrons}$ .

Formula used:

The mole of nikel is,

$\text{mole of nickel}=\frac{\text{mass of nickel}}{\text{mass of each mole of coin's atoms}}$

Calculation:

Consider 1 mole contains $6.022\times {10}^{23}\text{ atoms}$ .

The mole of nickel in a coin is,

$\begin{array}{l}\text{mole of nickel}=\frac{\text{5 g}}{\text{62 g}}\\ \text{mole of nickel}=0.08\text{ moles}\end{array}$

The number of atoms in a nickel is,

$\begin{array}{l}\text{number of atoms}=\left(0.08\text{ moles}\right)\times \frac{6.022\times {10}^{23}\text{ atoms}}{\text{1}\text{.0 moles}}\\ \text{number of atoms}=5\times {10}^{22}\text{ atoms}\end{array}$

The number of electrons in the coin for nickel is,

$\begin{array}{l}\text{number of electrons}=\left(5\times {10}^{22}\text{ atoms}\right)\times \frac{28\text{ electrons}}{1\text{ atom}}\\ \text{number of electrons}=1.4\times {10}^{24}\text{ electrons}\end{array}$

So, the coulombs of chargeon the electrons of a nickel is,

$\begin{array}{l}\text{coulombs on electrons}=\left(1.4\times {10}^{24}\text{ electrons}\right)\times \frac{\text{1}\text{.6}\times {\text{10}}^{-19}\text{ coulombs}}{1\text{ electron}}\\ \text{coulombs on electrons}=\text{2}\text{.24}\times {\text{10}}^5\text{ coulombs}\end{array}$

Conclusion:

Thus, the coulombs of charge on the electrons of a nickel is $\text{2}\text{.24}\times {\text{10}}^5\text{ coulombs}$ .