41. X rays from a copper X-ray tube (lambda =154pm) were diffracted at an angle of 14.22 degrees by a crystal of silicon . Assuming first-order diffraction (n=1 in the Bragg equation). what is the interplanar spacing in silicon? 42. The second-order diffraction (n=2) for a gold crystal is at an angle of 22.20^circ for X rays of 154 pm. What is the spacing between these crystal planes?

41. To find the interplanar spacing in silicon, we can use the Bragg equation:<br /><br />\[ n\lambda = 2d\sin(\theta) \]<br /><br />where:<br />- \( n \) is the order of diffraction (first-order diffraction, \( n = 1 \))<br />- \( \lambda \) is the wavelength of the X-rays (154 pm or \( 154 \times 10^{-12} \) m)<br />- \( d \) is the interplanar spacing<br />- \( \theta \) is the angle of diffraction (14.22 degrees)<br /><br />Rearranging the equation to solve for \( d \):<br /><br />\[ d = \frac{n\lambda}{2\sin(\theta)} \]<br /><br />Substituting the given values:<br /><br />\[ d = \frac{1 \times 154 \times 10^{-12}}{2 \times \sin(14.22^\circ)} \]<br /><br />\[ d \approx 3.54 \times 10^{-10} \, \text{m} \]<br /><br />So, the interplanar spacing in silicon is approximately \( 3.54 \times 10^{-10} \) meters.<br /><br />42. For the second-order diffraction (\( n = 2 \)) for a gold crystal, we can use the same Bragg equation:<br /><br />\[ 2\lambda = 2d\sin(\theta) \]<br /><br />where:<br />- \( \lambda \) is the wavelength of the X-rays (154 pm or \( 154 \times 10^{-12} \) m)<br />- \( d \) is the interplanar spacing<br />- \( \theta \) is the angle of diffraction (22.20 degrees)<br /><br />Rearranging the equation to solve for \( d \):<br /><br />\[ d = \frac{2\lambda}{2\sin(\theta)} \]<br /><br />Substituting the given values:<br /><br />\[ d = \frac{2 \times 154 \times 10^{-12}}{2 \times \sin(22.20^\circ)} \]<br /><br />\[ d \approx 3.54 \times 10^{-10} \, \text{m} \]<br /><br />So, the spacing between these crystal planes in gold is approximately \( 3.54 \times 10^{-10} \) meters.