QuestionJune 17, 2025

When an X-ray beam with a wavelength of 120 pm strikes the surface of a crystal with a lattice spacing of 165 nm, what is the maximum reflection angle with a value of n=1 32.6^circ 54.0^circ 21.3^circ 38.3^circ 18.4^circ

When an X-ray beam with a wavelength of 120 pm strikes the surface of a crystal with a lattice spacing of 165 nm, what is the maximum reflection angle with a value of n=1 32.6^circ 54.0^circ 21.3^circ 38.3^circ 18.4^circ
When an X-ray beam with a wavelength of 120 pm strikes the surface of a crystal with a lattice spacing of 165 nm, what is the maximum reflection angle with a value of n=1 32.6^circ 54.0^circ 21.3^circ 38.3^circ 18.4^circ

Solution
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Answer

None of the provided angles match the calculated maximum reflection angle. Explanation 1. Convert units Convert the wavelength from picometers to nanometers: 120 \text{ pm} = 0.12 \text{ nm}. 2. Apply Bragg's Law Use **Bragg's Law**: n\lambda = 2d\sin(\theta), where n=1, \lambda=0.12 \text{ nm}, and d=165 \text{ nm}. 3. Solve for \theta Rearrange to find \theta: \sin(\theta) = \frac{n\lambda}{2d} = \frac{0.12}{2 \times 165}. 4. Calculate \theta Compute \sin(\theta) = \frac{0.12}{330} = 0.0003636. Find \theta using \theta = \arcsin(0.0003636). 5. Compare with given angles Calculate \theta \approx 0.0208^{\circ}. Compare with given options.

Explanation

1. Convert units<br /> Convert the wavelength from picometers to nanometers: $120 \text{ pm} = 0.12 \text{ nm}$.<br /><br />2. Apply Bragg's Law<br /> Use **Bragg's Law**: $n\lambda = 2d\sin(\theta)$, where $n=1$, $\lambda=0.12 \text{ nm}$, and $d=165 \text{ nm}$.<br /><br />3. Solve for $\theta$<br /> Rearrange to find $\theta$: $\sin(\theta) = \frac{n\lambda}{2d} = \frac{0.12}{2 \times 165}$.<br /><br />4. Calculate $\theta$<br /> Compute $\sin(\theta) = \frac{0.12}{330} = 0.0003636$. Find $\theta$ using $\theta = \arcsin(0.0003636)$.<br /><br />5. Compare with given angles<br /> Calculate $\theta \approx 0.0208^{\circ}$. Compare with given options.
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