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

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

Answer 11

mit would kick you out of the school for doing what you are doing here.