Maximum Wavelength and Scattering Angle for Triangular Lattice

[GL_Black_Hole]

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.

 

[anathema]

To find the possible values of h and k, we can use the fact that the lattice spacing is 2 angstroms. This means that the possible values of h and k will be multiples of 2, since each unit cell in the lattice has a length of 2 angstroms.

So, we can start by finding the smallest possible values of h and k, which would be h=1 and k=1. Plugging these values into the equations for $\beta_{scat}$, we get:

$$k_0(cos\beta -cos\alpha) = \frac{2\pi}{a}$$
$$k_0(sin\beta -sin\alpha) = \frac{2\pi}{\sqrt{3}a}$$

We can rearrange these equations to solve for cosβ and sinβ:

$$cos\beta = \frac{k_0a}{2\pi}cos\alpha + \frac{1}{2}$$
$$sin\beta = \frac{k_0a}{2\pi}sin\alpha + \frac{1}{\sqrt{3}}$$

Since we know the values of α, a, and k0, we can plug them in to find the values of cosβ and sinβ. This will give us two possible solutions for β, since there are two possible values for h and k.

Once we have these two solutions, we can plug them back into the original equation for β and solve for the scattering angle Ω. This will give us two possible values for Ω, since there are two possible values for β.

We can then use these two values for Ω to find the corresponding values for βscat at λmax, using the equation:
$$\beta_{scat} = \frac{\Omega}{2}$$

So, in summary, to find the possible values of h and k, we can use the fact that the lattice spacing is 2 angstroms and plug in the smallest possible values of h and k (h=1, k=1) to solve for β. Then, we can plug these values into the equations for Ω and βscat to find the two possible values for each.