A copper atom consists of copper nucles surrounded by 29 electrons. The atomic weight of copper is `63.5 gmol^(-1)`. Let us now take two pieces of copper each weighing 10g. Let one electron from one piece be transferred to another for every 1000 atom in a piece. ltbRgt (a) Find the magnitude of charge appearing on each piece.
(b) What will be the Coulomb force between the two pieces after the transfer of electrons if they are 10cm apart?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as given in the question.

### Part (a): Find the magnitude of charge appearing on each piece.

1. **Calculate the number of atoms in 10 g of copper:**
\[ \text{Number of atoms} = \frac{\text{Weight of copper}}{\text{Atomic weight}} \times \text{Avogadro's number} \]
Given:
- Weight of copper = 10 g
- Atomic weight of copper = 63.5 g/mol
- Avogadro's number \( N_A = 6 \times 10^{23} \) atoms/mol

Substituting the values:
\[ n = \frac{10 \, \text{g}}{63.5 \, \text{g/mol}} \times 6 \times 10^{23} \, \text{atoms/mol} \]
\[ n \approx 9.448 \times 10^{22} \, \text{atoms} \]

2. **Calculate the number of electrons transferred:**
It is given that one electron is transferred for every 1000 atoms.
\[ \text{Number of electrons transferred} = \frac{n}{1000} \]
\[ N = \frac{9.448 \times 10^{22}}{1000} = 9.448 \times 10^{19} \, \text{electrons} \]

3. **Calculate the magnitude of charge on each piece:**
The charge \( Q \) is given by:
\[ Q = N \times e \]
Where \( e \) (the charge of one electron) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
\[ Q = 9.448 \times 10^{19} \times 1.6 \times 10^{-19} \]
\[ Q \approx 15.12 \, \text{C} \]

### Part (b): Calculate the Coulomb force between the two pieces after the transfer of electrons if they are 10 cm apart.

1. **Use Coulomb's Law to find the force:**
The formula for the Coulomb force \( F \) is:
\[ F = \frac{k \cdot Q^2}{R^2} \]
Where:
- \( k \) (Coulomb's constant) = \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( Q \) = charge on each piece = \( 15.12 \, \text{C} \)
- \( R \) = distance between the pieces = \( 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Substituting the values into the formula:**
\[ F = \frac{9 \times 10^9 \cdot (15.12)^2}{(0.1)^2} \]
\[ F = \frac{9 \times 10^9 \cdot 228.0644}{0.01} \]
\[ F = 2.05 \times 10^{14} \, \text{N} \]

### Final Answers:
(a) The magnitude of charge appearing on each piece is approximately **15.12 C**.
(b) The Coulomb force between the two pieces is approximately **\( 2.05 \times 10^{14} \, \text{N} \)**.