Question

A sample of an oriented semi-crystalline polymer is subjected to uniaxial tensile stress, \( \sigma \), in an X-ray diffractometer. The wavelength of X-ray radiation (Cu K\( \alpha \)) is \( \lambda = 1.542 \, \text{Å} \). The position of the (002) peak, which was found initially at a Bragg angle of 37.50° at \( \sigma = 0 \, \text{MPa} \), shifted to 37.45° at \( \sigma = 160 \, \text{MPa} \). Assuming elastic deformation, the strain (rounded off to three decimal places) in the sample along the direction of applied stress is \(\underline{\hspace{2cm}}\) \( \times 10^{-3} \).

Hint:

To calculate the strain in crystalline polymers under stress, use the shift in the Bragg angle and apply Bragg's law.

Answer 1

Correct Answer: 1

Using Bragg's law:
\[ n\lambda = 2d \sin \theta \] where \( n = 1 \) and \( \lambda = 1.542 \, \text{Å} \).
The change in the Bragg angle \( \Delta \theta = 37.50^\circ - 37.45^\circ = 0.05^\circ \).
The strain \( \epsilon \) is given by:
\[ \epsilon = \frac{\Delta \theta}{\theta_0} \] Substitute the values:
\[ \epsilon = \frac{0.05}{37.50} = 0.00133. \] Thus, the strain is approximately \( 1.33 \times 10^{-3} \).