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^(9) N m^(-2)`
D. `2.33xx10^(10) N m^(-2)`

Answer

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) is given by:

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

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 division:**
- Calculate \( \frac{7}{3} \):
\[ \frac{7}{3} \approx 2.3333 \]
- Now, divide by \( 10^{-10} \):
\[ Y \approx 2.3333 \times 10^{10} \, \text{N/m}^2 \]

4. **Round the result to two decimal places:**
\[ Y \approx 2.33 \times 10^{10} \, \text{N/m}^2 \]

### 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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