- #1

GL_Black_Hole

- 21

- 0

## Homework Statement

A triangular lattice of lattice spacing ##a=2 ## angstroms is irradiated with x-rays at time zero of wavelength 20 angstroms at an incident angle of ##\alpha =135##.

1) What is the maximum wavelength of the incident x-rays?

2) What is the scattering angle ##\Omega## for observed radiation?

3) What is ##\beta_{scat}## at ##\lambda_{max}##? Where ##\beta_{scat}## is the the angle of the scattered radiation relative to the x-direction of the lab frame.

## Homework Equations

##\Delta k = G## , ##G = hb_1 +kb_2##, ## k =|k| =2\pi/\lambda##, ##b_1 = 2\pi/V_{cell} (a_2 \times a_3)##

## The Attempt at a Solution

1) From ##\Delta k=G## we can take magnitudes of both sides and get: ##2k^2 (1-cos(\Omega)) =|G|^2##, or calling ##\Omega =2\theta##, we get ##4k^2 sin^2(\theta) =|G|^2##. The maximum value of this expression is when ##k =|G|/2##. So the max wavelength is ##4\pi/|G|##.

The lattice translation vectors are: ##a_1 = a \hat{x}##, ##a_2 = a/2 \hat{x} +\frac{\sqrt{3}}{2} a\hat{y}##, so the reciprocal lattice vectors are: ##b_1 = \frac{2\pi}{a} \hat{x} -\frac{2\pi}{\sqrt{3}a}\hat{y}##, ##b_2 =\frac{4\pi}{\sqrt{3}a}\hat{y}##. This implies that ##|G|^2 = \frac{16\pi^2}{3a^2} (h^2 +k^2)##, so ##\lambda_{max} =\frac{\sqrt{3}a}{\sqrt{h^2+k^2}}##.

2) The scattering angle ##\Omega## is determined by ##2k_0^2(1-cos\Omega) = \frac{16\pi^2}{3a^2} (h^2+k^2)##.

3) ##\beta_{scat}## is determined by the system of equations:

##k_0(cos\beta -cos\alpha) = h\frac{2\pi}{a}, k_0(sin\beta -sin\alpha) = \frac{4\pi k}{\sqrt{3}a} -\frac{2\pi h}{\sqrt{3}a}##, and so at max wavelength, the smallest possible value of ##h^2 +k^2##, we can solve for ##\beta##. I'm just unsure of how to find to the possible values of h,k.