# Maximum Wavelength and Scattering Angle for Triangular Lattice

• GL_Black_Hole
In summary: These values will give us the maximum scattering angle and the angle of the scattered radiation relative to the x-direction of the lab frame at the maximum wavelength of the incident x-rays.
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.

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.

## 1. What is the maximum wavelength for a triangular lattice?

The maximum wavelength for a triangular lattice is equal to twice the distance between lattice points. This is known as the lattice constant and is denoted by "a".

## 2. How is the maximum wavelength calculated for a triangular lattice?

The maximum wavelength can be calculated using the formula λ_max = 2a, where "a" is the lattice constant. This formula is derived from the Bragg's law, which states that the maximum wavelength is twice the distance between lattice planes.

## 3. What is the significance of the maximum wavelength in a triangular lattice?

The maximum wavelength determines the range of wavelengths that can be diffracted by a triangular lattice. Wavelengths longer than the maximum will not be diffracted, while shorter wavelengths will be diffracted at various angles.

## 4. Can the maximum wavelength be altered in a triangular lattice?

Yes, the maximum wavelength can be altered by changing the lattice constant "a". This can be done by changing the size or composition of the lattice, or by applying external forces such as strain.

## 5. What is the scattering angle for a maximum wavelength in a triangular lattice?

The scattering angle for a maximum wavelength in a triangular lattice is 90 degrees. This means that the diffracted beam will be perpendicular to the incident beam, resulting in a bright spot in the diffraction pattern.

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