X-rays with a wavelength of 1.49 Å scatter at an angle of 14.0 from a crystal. If n1 what is the distance between planes of atoms in the crystal that give rise to this scattering?

To find the distance between the planes of atoms in the crystal, we can use Bragg's law, which is given by the equation ( n\lambda = 2d\sin(\theta) ). Here, ( \lambda ) is the wavelength (1.49 Å), ( \theta ) is the scattering angle (14.0°), and ( n ) is the order of reflection (which you have noted as ( n = 1 )). Rearranging the equation to solve for ( d ) gives:

[ d = \frac{n\lambda}{2\sin(\theta)} = \frac{1 \times 1.49 , \text{Å}}{2\sin(14.0°)} \approx 5.17 , \text{Å}. ]

Thus, the distance between the planes of atoms in the crystal is approximately 5.17 Å.