Question

The young's modulus for steal is 2*10^11 N/M^2. If the inter atomic spacing for the metal is 2.8 angstrom. Find the increase in the inter atomic spacing for a force of 10^9 N/M^2,and the force constant?

Answer 1

Answer on Question#57182 - Physics - Solid State Physics

The young's modulus for steal is $E = 2 * 10^{11} \frac{\mathrm{N}}{\mathrm{m}^2}$. If the inter atomic spacing for the metal is $a = 2.8$ angstrom. Find the increase in the inter atomic spacing $\Delta a$ for a force of $\sigma = 10^9 \frac{\mathrm{N}}{\mathrm{m}^2}$, and the force constant $k$?

Solution:

It's known that

$$\sigma = E \cdot \frac{\Delta a}{a}$$

Thus

$$\Delta a = a \frac{\sigma}{E} = 2.8 \, \text{\AA} \frac{10^9 \frac{\mathrm{N}}{\mathrm{m}^2}}{2 * 10^{11} \frac{\mathrm{N}}{\mathrm{m}^2}} = 0.014 \, \text{\AA}$$

Force constant:

$$k = \frac{E}{a} \cdot 1 \, \mathrm{m}^2 = \frac{2 * 10^{11} \frac{\mathrm{N}}{\mathrm{m}^2}}{2.8 \, \text{\AA}} \cdot 1 \, \mathrm{m}^2 = 7.14 \times 10^{20} \frac{\mathrm{N}}{\mathrm{m}}$$

Answer: $\Delta a = 0.014 \, \text{\AA}, k = 7.14 \times 10^{20} \frac{\mathrm{N}}{\mathrm{m}}$.

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