a)Calculate the lattice constant a in palladium (Pd).express your answer in angstroms.

b) Calculate the distance (in angstroms) between adjacent (110) planes in palladium (Pd).
Hint: use your work from part (a).

3.89

(a) is right and the (b)?

I can help you through this if you explain exactly what your trouble is.

The (b)?

(a) 3.89

(b) 2.75

thanks

To calculate the lattice constant of palladium (Pd), we need to reference the crystal structure of Pd. Palladium has a face-centered cubic (FCC) structure, which means each atom is surrounded by 12 nearest neighbors at the corners of a cube, in addition to 6 nearest neighbors on the faces of the cube.

a) To find the lattice constant (denoted by a), we can utilize the relationship between the atomic radius and the lattice constant in FCC structures. In an FCC structure, the length of the diagonal of a face of the unit cell is equal to 4 times the atomic radius (r).

Therefore, the lattice constant (a) can be found through the formula:
a = 4r / √2

To obtain the atomic radius, we can refer to reliable sources such as the "CRC Handbook of Chemistry and Physics." According to this source, the atomic radius of palladium is approximately 137 pm (picometers) or 1.37 angstroms.

Considering the given value of r = 1.37 angstroms, we can calculate the lattice constant as follows:
a = 4 * 1.37 / √2

Now, let's compute this value:
a ≈ 3.89 angstroms

Therefore, the lattice constant of palladium (Pd) is approximately 3.89 angstroms.

b) To determine the distance between adjacent (110) planes (d110) in palladium, we need to utilize the formula:
d110 = a / √(h^2 + k^2 + l^2)

In this case, for the (110) planes, h = 1, k = 1, and l = 0.

Substituting the given values and the lattice constant (a) we found in part (a), we can calculate the distance between adjacent (110) planes:

d110 = 3.89 / √(1^2 + 1^2 + 0^2)

Now, let's compute this value:
d110 ≈ 2.75 angstroms

Therefore, the distance between adjacent (110) planes in palladium (Pd) is approximately 2.75 angstroms.