The image contains a question related to X-ray diffraction on palladium with the following information:

• X-ray diffraction from the (111) plane of palladium is observed at a scattering angle of 39.30°.
• The X-ray wavelength is 0.154 nm.
• Palladium has an FCC (face-centered cubic) crystal structure.

The task asks to find the atomic radius of palladium and to express the answer using three significant figures.

Step-by-step approach to solving the problem:

1. Bragg's Law: Bragg's Law helps us relate the scattering angle to the lattice spacing, and it is given by: $n\lambda = 2d\sin\theta$ Where:

   • $n$ is the order of diffraction (usually $n = 1$),
   • $\lambda = 0.154 \, \text{nm}$ is the wavelength,
   • $\theta = 39.30^\circ$ is the scattering angle,
   • $d$ is the distance between the planes (lattice spacing).

2. Lattice spacing ($d$) can be calculated using Bragg’s Law for the (111) plane.

3. Lattice constant ($a$) for an FCC structure can be related to the interplanar spacing: $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$ For the (111) plane, $h = 1$, $k = 1$, and $l = 1$: $d_{111} = \frac{a}{\sqrt{3}}$

4. Atomic radius ($r$) for an FCC structure is related to the lattice constant $a$ by: $r = \frac{a}{2\sqrt{2}}$

Let me calculate the atomic radius of palladium using the above relations.The atomic radius of palladium, calculated using the given data and Bragg's Law, is approximately 0.074 nm when expressed with three significant figures.

Would you like more details, or do you have any further questions?

Related Questions:

1. How does Bragg’s Law apply to other crystal structures besides FCC?
2. What role does the wavelength of X-rays play in diffraction experiments?
3. How can one experimentally determine the scattering angle in an X-ray diffraction setup?
4. What is the significance of the (111) plane in face-centered cubic (FCC) structures?
5. How does the atomic radius relate to the density of a material?

Tip:

In X-ray diffraction, increasing the X-ray wavelength will generally reduce the scattering angle for a given plane, making it easier to measure precise angles.