(a) Calculate the lattice constant a in palladium (Pd). Use only the information available in the class Periodic Table, and 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).

a)3.89

b)2.75

To calculate the lattice constant (a) in palladium (Pd) using only the information available in the class Periodic Table, we need to find the atomic radius of palladium.

(a) Steps to calculate the lattice constant, a:

1. Find the atomic number (Z) of palladium on the periodic table. Palladium has an atomic number of 46.

2. Locate the atomic radius (r) for palladium. The atomic radius is typically half the distance between the nuclei of two adjacent atoms in a solid crystal lattice.

It's important to note that the atomic radius varies depending on the crystal structure. In this case, we'll consider palladium to have a face-centered cubic (FCC) crystal structure.

The FCC crystal structure consists of atoms located at the corners of a cube, with additional atoms at the centers of each face. The lattice constant (a) is the length of one side of the cube.

3. The FCC crystal structure involves a diagonal relationship between the atoms. The distance (d) between adjacent atoms along one of the cube's diagonals is related to the lattice constant (a) by the equation:
d = a * sqrt(2)
(where sqrt is the square root function)

4. Since we're looking for the lattice constant (a), we can rearrange the equation to solve for a:
a = d / sqrt(2)

5. Plug in the value of d, which is the atomic radius (r) of palladium, divided by sqrt(2):
a = r / sqrt(2)

6. Look up the atomic radius of palladium on the periodic table. The atomic radius of palladium is approximately 1.39 angstroms (Å).

7. Substitute the atomic radius into the equation and solve:
a = 1.39 Å / sqrt(2)
a ≈ 0.9835 Å

Therefore, the lattice constant (a) in palladium is approximately 0.9835 angstroms (Å).

(b) To calculate the distance between adjacent (110) planes in palladium using the result from part (a):

1. The (110) planes refer to a specific set of crystallographic planes in the FCC structure.

2. The distance (d) between adjacent (110) planes can be calculated using the relationship:
d = a / sqrt(h^2 + k^2 + l^2)
where h, k, and l are the Miller indices representing the specific crystallographic plane.

3. For the (110) planes, h = 1, k = 1, and l = 0.

Substituting these values, and using the value of a obtained from part (a):
d = 0.9835 Å / sqrt(1^2 + 1^2 + 0^2)
d = 0.9835 Å / sqrt(2)
d ≈ 0.6969 Å

Therefore, the distance between adjacent (110) planes in palladium is approximately 0.6969 angstroms (Å).