Two copper balls, each weighing `10 g` are kept in air `10 cm` apart. If one electron from every `10^(6)` atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is `63.5`)

Options

A. `2.0xx10^(10)N`
B. `2.0xx10^(4)N`
C. `2.0xx10^(8)N`
D. `2.0xx10^(6)N`

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Determine the number of moles of copper in each ball.
Given:
- Weight of each copper ball = 10 g
- Atomic weight of copper = 63.5 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
For one ball:
\[ \text{Number of moles} = \frac{10 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.1575 \, \text{mol} \]

### Step 2: Calculate the total number of atoms in one ball.
Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol):
\[ \text{Total number of atoms} = \text{Number of moles} \times \text{Avogadro's number} \]
\[ = 0.1575 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 9.46 \times 10^{22} \, \text{atoms} \]

### Step 3: Determine the number of electrons transferred.
We know that one electron is transferred from every \(10^6\) atoms. Thus, the number of electrons transferred from one ball to the other is:
\[ \text{Number of electrons transferred} = \frac{9.46 \times 10^{22} \, \text{atoms}}{10^6} \approx 9.46 \times 10^{16} \, \text{electrons} \]

### Step 4: Calculate the total charge transferred.
The charge of one electron is approximately \(1.6 \times 10^{-19}\) C. Therefore, the total charge (\(Q\)) transferred is:
\[ Q = \text{Number of electrons} \times \text{Charge of one electron} \]
\[ Q = 9.46 \times 10^{16} \times 1.6 \times 10^{-19} \approx 1.5136 \times 10^{-2} \, \text{C} \]

### Step 5: Calculate the Coulomb force between the two balls.
Using Coulomb's law:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
Where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
- \(Q_1 = Q_2 = 1.5136 \times 10^{-2} \, \text{C}\)
- \(r = 10 \, \text{cm} = 0.1 \, \text{m}\)

Substituting the values:
\[ F = 9 \times 10^9 \frac{(1.5136 \times 10^{-2})^2}{(0.1)^2} \]
Calculating:
\[ F = 9 \times 10^9 \frac{(2.2953 \times 10^{-4})}{0.01} = 9 \times 10^9 \times 2.2953 \times 10^{-2} \]
\[ F \approx 2.0658 \times 10^8 \, \text{N} \]

### Final Answer:
The Coulomb force between the two copper balls is approximately \(2.07 \times 10^8 \, \text{N}\).

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