Calculate the surface energy of the (100) surface of niobium. The enthalpy of atomization of niobium is 745 kJ/mole. Express your answer in units of J/cm2.
Could someone insert the formula and data, at least it gives an idea in how-to answer the question?
9.09e-3 J/cm2
wrong
2.8*10^-4
the formula pleases!
it's wrong
cheat 3.091 x 10^-6 joules/cm2
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
Dr B, is wrong
6.23 e -3, works. not sure why though
beware. course staff are flagging students who are entering planted answers.
Marcia 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)