What is the distance between second nearest neighbor atoms in vanadium? Express your answer in cm
To determine the distance between second nearest neighbor atoms in vanadium, we need to know the crystal structure of vanadium. Vanadium has a body-centered cubic (BCC) crystal structure.
In a BCC crystal structure, the second nearest neighbor atoms are located along the body diagonal of the cube. The body diagonal is calculated using the formula:
d = a√3
where d is the length of the body diagonal and a is the side length of the cube.
For vanadium, the lattice constant (a) is approximately 3.03 Å (angstroms, which is 1 × 10^-8 cm).
To convert the lattice constant to centimeters, we can multiply it by the conversion factor: 1 Å = 1 × 10^-8 cm.
So, the lattice constant in centimeters is 3.03 × 10^-8 cm.
Now, we can calculate the distance between second nearest neighbor atoms:
d = a√3 = 3.03 × 10^-8 cm × √3 = 3.03 × √3 × 10^-8 cm
Calculating this expression gives us:
d ≈ 1.66 × 10^-7 cm
Therefore, the distance between second nearest neighbor atoms in vanadium is approximately 1.66 × 10^-7 cm.