An X-ray beam of wavelength 3.4 x 10-10 m makes an angle of 27° with a set of planes in a crystal which results in first order constructive interference. Determine the plane spacing in nanometers. (Please include 2 decimal places).

Physics for Scientists and Engineers
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Author:Raymond A. Serway, John W. Jewett
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Chapter36: Wave Optics
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### X-ray Diffraction Exercise

**Problem Statement:**

An X-ray beam of wavelength \( 3.4 \times 10^{-10} \) meters makes an angle of \( 27^\circ \) with a set of planes in a crystal which results in first order constructive interference. Determine the plane spacing in nanometers. (Please include 2 decimal places).

**Solution:**
To solve this problem, you will need to use Bragg's Law which is represented by the equation:

\[ n\lambda = 2d\sin\theta \]

where:
- \( n \) is the order of the interference (for first order, \( n = 1 \))
- \( \lambda \) is the wavelength of the X-ray ( \( 3.4 \times 10^{-10} \) meters)
- \( d \) is the distance between the crystal planes
- \( \theta \) is the angle of incidence ( \( 27^\circ \) )

Rearranging the formula to solve for \( d \):

\[ d = \frac{n\lambda}{2\sin\theta} \]

Step-by-step calculation:

1. Substitute the known values into the equation:

\[ d = \frac{(1)(3.4 \times 10^{-10} \, \text{m})}{2\sin(27^\circ)} \]

2. Calculate \( \sin(27^\circ) \). Use a calculator to find:

\[ \sin(27^\circ) \approx 0.454 \]

3. Substitute this value back into the equation:

\[ d = \frac{3.4 \times 10^{-10}}{2 \times 0.454} \]

4. Perform the division:

\[ d \approx \frac{3.4 \times 10^{-10}}{0.908} \approx 3.74 \times 10^{-10} \, \text{m} \]

5. Convert the result from meters to nanometers. (1 \, \text{m} = 10^9 \, \text{nm}):
   
\[ d \approx 3.74 \times 10^{-1} \, \text{nm} \]

Thus, the plane spacing \( d \) is approximately \( 0.37 \, \text{nm} \).

**Conclusion:**
The plane spacing is \(
Transcribed Image Text:### X-ray Diffraction Exercise **Problem Statement:** An X-ray beam of wavelength \( 3.4 \times 10^{-10} \) meters makes an angle of \( 27^\circ \) with a set of planes in a crystal which results in first order constructive interference. Determine the plane spacing in nanometers. (Please include 2 decimal places). **Solution:** To solve this problem, you will need to use Bragg's Law which is represented by the equation: \[ n\lambda = 2d\sin\theta \] where: - \( n \) is the order of the interference (for first order, \( n = 1 \)) - \( \lambda \) is the wavelength of the X-ray ( \( 3.4 \times 10^{-10} \) meters) - \( d \) is the distance between the crystal planes - \( \theta \) is the angle of incidence ( \( 27^\circ \) ) Rearranging the formula to solve for \( d \): \[ d = \frac{n\lambda}{2\sin\theta} \] Step-by-step calculation: 1. Substitute the known values into the equation: \[ d = \frac{(1)(3.4 \times 10^{-10} \, \text{m})}{2\sin(27^\circ)} \] 2. Calculate \( \sin(27^\circ) \). Use a calculator to find: \[ \sin(27^\circ) \approx 0.454 \] 3. Substitute this value back into the equation: \[ d = \frac{3.4 \times 10^{-10}}{2 \times 0.454} \] 4. Perform the division: \[ d \approx \frac{3.4 \times 10^{-10}}{0.908} \approx 3.74 \times 10^{-10} \, \text{m} \] 5. Convert the result from meters to nanometers. (1 \, \text{m} = 10^9 \, \text{nm}): \[ d \approx 3.74 \times 10^{-1} \, \text{nm} \] Thus, the plane spacing \( d \) is approximately \( 0.37 \, \text{nm} \). **Conclusion:** The plane spacing is \(
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