If the interatomic spacing in a steel wire is 3.0Å and Yₛₜₑₑₗ = 20 x10¹⁰ N / m² , then force constant is

The correct option is 6 x 10⁻⁹ N /Å.

Step-by-step derivation:

The relationship between the Young's modulus of a wire ($Y$), the interatomic spacing ($r_0$), and the interatomic force constant ($k$) is given by the formula:
$k=Y\times r_0$

We are given the following values:
• Young's modulus of steel, $Y_{\text{steel}}=20\times {10}^{10}{\text{N/m}}^2$
• Interatomic spacing, $r_0=3.0\text{Å}=3.0\times {10}^{-10}\text{m}$

First, calculate the force constant $k$ in SI units ($\text{N/m}$):
$k=(20\times {10}^{10}{\text{N/m}}^2)\times (3.0\times {10}^{-10}\text{m})$
$k=60\text{N/m}$

Now, convert the force constant into the unit $\text{N/Å}$. Since $1\text{m}={10}^{10}\text{Å}$, we have:
$k=\frac{60\text{N}}{{10}^{10}\text{Å}}$
$k=60\times {10}^{-10}\text{N/Å}$
$k=6\times {10}^{-9}\text{N/Å}$