The mean distance between the atoms of iron is `3xx10^(-10)` m and interatomic fore constant for iron is `7 N//m`. The Young's modulus of elasticity for iron is

Options

A. `2.33xx10^(5) N//m^(2)`
B. `23.3xx10^(10)N//m^(2)`
C. `2.33xx10^(10)N//m^(2)`
D. `2.33xx10^(12)N//m^(2)`

Explanation

To find the Young's modulus of elasticity for iron, we can use the relationship between the interatomic force constant and the mean distance between the atoms. The formula for Young's modulus (Y) can be expressed as:

\[ Y = \frac{k}{d} \]

where:
- \( k \) is the interatomic force constant,
- \( d \) is the mean distance between the atoms.

### Step-by-step solution:

1. **Identify the given values:**
- Mean distance between the atoms of iron, \( d = 3 \times 10^{-10} \) m
- Interatomic force constant for iron, \( k = 7 \, \text{N/m} \)

2. **Substitute the values into the Young's modulus formula:**
\[ Y = \frac{k}{d} = \frac{7 \, \text{N/m}}{3 \times 10^{-10} \, \text{m}} \]

3. **Perform the calculation:**
\[ Y = \frac{7}{3 \times 10^{-10}} = \frac{7}{3} \times 10^{10} \, \text{N/m}^2 \]

4. **Calculate \( \frac{7}{3} \):**
\[ \frac{7}{3} \approx 2.33 \]

5. **Combine the results:**
\[ Y \approx 2.33 \times 10^{10} \, \text{N/m}^2 \]

### Final Answer:
The Young's modulus of elasticity for iron is approximately \( 2.33 \times 10^{10} \, \text{N/m}^2 \).

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