Estimate the surface energy of (110) face of K crystals given just the information in the periodic table. Express your answer in the unit of J/cm2.

Someone MIGHT be able to help you IF you indicate what YOU THINK about each of these problems you've posted.

First Consider the Lattice Type of Potassium , and that is BBC

Then determine where the 110 "plane" is

Now, if you can imagine the (110) position you can see that cutting the unit cell at that position will rapture two K-K bonds. Hence to know that surface energy you need to know what is the value of breaking a K-K bond!!! then divide it by the area of that (110) surface.

To know K-K bond energy, you can start with (Ha) Atomisation energy of K given in the periodic table:

89.54 kJ/mol.

Ok you can convert that per atom value

So divide it by Avogadros number

89.54 KJ/mol x (1 mol / 6.02E23)

Now look at the BCC structure again and you see that each Na atom has 8 atomic neighbors, hence for one K atom to atomize it need to break 8 bonds, so now K-K bond energy = 89.54 KJ/mol x (1 mol / 6.02E23) / 8 !!!

Now that you have that take note that the unit is KJ/bond now. You must now calculate the area of 110 plan. To do that you have to solve for a (Lattice Constant)

a = (2 * Molar Volume / Avogadros) ^ 1/3

Now the area of the plane is just a*a^(1/2)

You will then notice that your final units is KJ/(bond cm^2)

You need to remove that bond unit so you must divide your final answer with the number of bonds present at the surface you wish to break / slip, and again that number of bonds is 2 !!!

To estimate the surface energy of the (110) face of K crystals using just the information in the periodic table, we can make use of the relationship between surface energy and interatomic spacing.

The surface energy of a crystal is primarily determined by the bonding forces between atoms on the surface. In general, the surface energy decreases as the interatomic spacing increases because the bonding forces become weaker.

To estimate the surface energy, we can consider the atomic radius of potassium (K), which can be found on the periodic table. The atomic radius provides an estimate of the interatomic spacing in the crystal structure.

The atomic radius of K is approximately 2.31 Å (angstroms), which is equivalent to 2.31 x 10^-8 cm. Since we are interested in the (110) face, which is a specific crystallographic plane, we need to consider the number of atoms per unit area.

For the (110) face of a simple cubic crystal structure, there are 2 atoms per unit area. Therefore, the interatomic spacing is given by half the atomic radius, which is 1.155 x 10^-8 cm.

Now, we can estimate the surface energy using the equation:

Surface Energy = γ = (E/2A)

Where γ is the surface energy, E is the total energy of the surface, and A is the surface area.

Since we don't have the necessary information to determine the total energy or surface area, we cannot directly calculate the surface energy of the (110) face of K crystals. Unfortunately, estimating the surface energy solely based on the periodic table is not possible without additional data or calculations.

It's worth mentioning that accurate surface energy calculations usually require more detailed information about the crystal structure, such as lattice parameters and interatomic potentials, which are beyond the scope of the periodic table alone.