Two copper pieces, each of mass 0.0635 kg are placed at a distance of 0.1 m from each other One electron from each atom of piece 1 is transferred to piece 2 of copper. Net charges on piece 1 and piece 2 after transfer of electrons respectively will be

Options

A. `q_1=-1.6 x 10^(-19) C,q_2= +1.6 x 10^(-19)C`
B. `q_1=-9.6 x 10^(4) C,q_2= +9.6 x 10^(4)C`
C. `q_1=+1.6 x 10^(-19) C,q_2= -1.6 x 10^(-19)C`
D. `q_1=+9.6 x 10^(4) C,q_2= -9.6 x 10^(4)C`

Answer

To solve the problem, we need to calculate the net charges on two pieces of copper after transferring one electron from each atom of piece 1 to piece 2.

### Step-by-Step Solution:

1. **Identify the Mass of Each Copper Piece**:
Each copper piece has a mass of \(0.0635 \, \text{kg}\).

2. **Convert Mass to Grams**:
\[ 0.0635 \, \text{kg} = 63.5 \, \text{g} \]

3. **Determine the Molar Mass of Copper**:
The molar mass of copper (Cu) is \(63.5 \, \text{g/mol}\).

4. **Calculate the Number of Moles in Each Copper Piece**:
Using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
For each piece:
\[ n_1 = n_2 = \frac{63.5 \, \text{g}}{63.5 \, \text{g/mol}} = 1 \, \text{mol} \]

5. **Calculate the Number of Atoms in Each Copper Piece**:
Using Avogadro's number (\(N_A = 6.02 \times 10^{23} \, \text{atoms/mol}\)):
\[ \text{Number of atoms} = n \times N_A \]
For each piece:
\[ \text{Number of atoms} = 1 \, \text{mol} \times 6.02 \times 10^{23} \, \text{atoms/mol} = 6.02 \times 10^{23} \, \text{atoms} \]

6. **Determine the Charge of One Electron**:
The charge of one electron is:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]

7. **Calculate the Charge on Piece 1 After Electron Transfer**:
When one electron is transferred from each atom of piece 1 to piece 2, piece 1 loses electrons. Thus, the charge on piece 1 becomes positive:
\[ \text{Total electrons removed} = 6.02 \times 10^{23} \, \text{electrons} \]
Therefore, the charge on piece 1:
\[ Q_1 = \text{Number of protons left} \times e = 6.02 \times 10^{23} \times 1.6 \times 10^{-19} \, \text{C} = 9.632 \times 10^{4} \, \text{C} \]
Hence, the charge on piece 1 is:
\[ Q_1 = +9.632 \times 10^{4} \, \text{C} \]

8. **Calculate the Charge on Piece 2 After Electron Transfer**:
Piece 2 gains the same number of electrons, thus it becomes negatively charged:
\[ Q_2 = -6.02 \times 10^{23} \times 1.6 \times 10^{-19} \, \text{C} = -9.632 \times 10^{4} \, \text{C} \]

9. **Final Charges**:
The net charges on piece 1 and piece 2 after the transfer of electrons are:
\[ Q_1 = +9.632 \times 10^{4} \, \text{C} \quad \text{and} \quad Q_2 = -9.632 \times 10^{4} \, \text{C} \]

### Summary of Results:
- Net charge on piece 1: \( +9.632 \times 10^{4} \, \text{C} \)
- Net charge on piece 2: \( -9.632 \times 10^{4} \, \text{C} \)