If the interatomic spacing in a steel wire is `3.0 Å and Y_(steel)=20xx10^(10)N//m^(2)` then force constant is

Options

A. `6xx10^(-2)N//Å`
B. `6xx10^(-9)N//Å`
C. `4xx10^(-5)N//Å`
D. `6xx10^(-5)N//Å`

Answer

To find the force constant of a steel wire given the interatomic spacing and Young's modulus, we can use the following relationship:

\[ k = Y \cdot d \]

where:
- \( k \) is the force constant,
- \( Y \) is Young's modulus,
- \( d \) is the interatomic spacing.

### Step-by-Step Solution:

1. **Identify the given values**:
- Interatomic spacing \( d = 3.0 \, \text{Å} = 3.0 \times 10^{-10} \, \text{m} \)
- Young's modulus for steel \( Y = 20 \times 10^{10} \, \text{N/m}^2 \)

2. **Convert interatomic spacing to meters**:
- Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we convert \( 3.0 \, \text{Å} \):
\[ d = 3.0 \, \text{Å} = 3.0 \times 10^{-10} \, \text{m} \]

3. **Substitute the values into the formula**:
\[ k = Y \cdot d = (20 \times 10^{10} \, \text{N/m}^2) \cdot (3.0 \times 10^{-10} \, \text{m}) \]

4. **Calculate the force constant**:
\[ k = 20 \times 3.0 \times 10^{10} \times 10^{-10} \, \text{N/m} \]
\[ k = 60 \, \text{N/m} \]

5. **Express the force constant in terms of angstroms**:
- Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we can express \( k \) in terms of angstroms:
\[ k = 60 \, \text{N/m} = 60 \times 10^{10} \, \text{N/Å} \]

6. **Final conversion**:
\[ k = 6 \times 10^{11} \, \text{N/Å} \]

### Final Answer:
The force constant \( k \) is \( 6 \times 10^{11} \, \text{N/Å} \).

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