calculate the surface energy of 100 surface of niobium. The enthalpy of atomization of niobium is 745 kj/mol. express anser inunits of j/cm^2

Answer 1

2.8*10^-4

Answer 2

It's wrong btw

Answer 3

The enthalpy of atomization of nobium is 745 KJ/mole.

The energy per atom is = 745000/Avogadro Number = 1.2371305e-18 J/atom

In a BCC structure each atom has 8 neighbors. Then
The energy per bond is = 1.2371305e-18 / 8 = 1.5464132e-19 J/bond

Now in a (100) plane the atoms density per unit area is = 4 (1/4) / a * a = 1/a2

For Nobium a = (2*92.91 / 8.57 * Avogadro number)^(1/3) = 3.30210056e-8 cm

Then the atoms density is 9.1710574e+14 atoms/cm2

Finally:The broken bonds are 4 per UC then (half for each surface)
Surface energy (100) =1/2 * 4 * Atoms density * Bond Energy = 0.00028364488 J/cm2

Answer 4

so wrong

Answer 5

try ivos idea