Question

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

Express your answer in J/cm2.

Answer 1

3.091 cheat

Answer 2

manfred, your answer is WRONG

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

5.6542e-4 J/cm2

Answer 5

It's wrong

Answer 6

this is an mit 3.091 cheat.

Answer 7

got it. drb is a dink.

6.53 e -3, works. not sure why though

Answer 8

this is an exam question

cheater alert

Answer 9

its wrong

Answer 10

So what the heck you all are doing here who says it is cheating