In an x-ray diffraction experiment,using x-rays of wavelength λ = 0.434 × 10−¹⁰ m,a first-order maximum occurred at 8° off the crystal plane.Find the distance d between crystal planes. A)1.56E-10 m B)2.19E-11 m C)4.88E-10 m D)1.70E-9 m E)3.12E-10 m

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The distance between crystal planes, $d$, can be found using Bragg's law, $n\lambda = 2d\sin(\theta)$, where $n$ is the order of the maximum, $\lambda$ is the wavelength of the X-rays, and $\theta$ is the angle from the crystal plane. Given $n = 1$, $\lambda = 0.434 \times 10^{-10}$ m, and $\theta = 8^\circ$, we can solve for $d$. Rearranging Bragg's law gives $d = \frac{\lambda}{2\sin(\theta)}$. Substituting the given values and calculating gives $d \approx 1.56 \times 10^{-10}$ m.

In an X-Ray Diffraction Experiment,using X-Rays of Wavelength λ =