Asked by nonu

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

Express your answer in J/cm2.
Answered by manfred
3.091 cheat
Answered by marcia
manfred, your answer is WRONG
Answered by DrB
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
Answered by xxxx
5.6542e-4 J/cm2
Answered by Anonymous
It's wrong
Answered by blotto largo
this is an mit 3.091 cheat.
Answered by ivo
got it. drb is a dink.

6.53 e -3, works. not sure why though
Answered by get a life, you idiot
this is an exam question

cheater alert
Answered by Anonymous
its wrong
Answered by Anonymous
So what the heck you all are doing here who says it is cheating
Answered by ethics levels have serious dropped
mit would kick you out of the school for doing what you are doing here.
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