Calculate the surface energy of the (100) plane of niobium. The heat of atomization is 745 kJ/mol.

Question

Express your answer in J/cm2.

Answer 1

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 (consider half for each surface)
Surface energy (100) =1/2 * 4 * Atoms density * Bond Energy = 0.00028364488 J/cm2

Answer 2

incorrect because bond energy is off by a factor of two because at 8 per atom you counted bonds twice.

and this is an mit 3.091 cheat. shame on you.

Answer 3

ivo said 6.23 e -3 works. anyone try?

Answer 4

yeb..wrong again

Answer 5

6.23 e -3 is wrong..

Answer 6

this is an exam question

cheater alert