Calculate the surface energy of the (100) plane of niobium. The heat of atomization is 745 kJ/mol.
Express your answer in J/cm2.
3.091 E -5 joules / square centimeter
I put the answer of ongrwa and it's WRONG
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
it's wrong. why is it wrong?
incorrect because bond energy is off by a factor of two because at 8 per atom you count bonds twice.
and this is an mit 3.091 cheat. shame on you.
6.32 e -3, works. not sure why though
the internet knows more about what you are doing than you know yourself!
I don't know why it is wrong. But I'm sure about the answer.
The atoms you need broke on (100) plane are the four in the plane bonded to the body central atom (remember that each atom in a BCC Crystal has 8 neighbors but you only need broke 4 - if you broke the 8 bonds you atomize the crystal)
why you keep insisting? can you not read?
also, this is an exam question
cheater alert
boot this guy off the list
I think asking a question when you don't know how to solve it is the best thing one could do. Also, nobody should point fingers when somebody is cheating. It's their problem, not yours.
About the question...I can't solve it either. I looked for explanations. Couldn't find them.