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View Full Version : solid state - crystallisation energy, stuck please help!


kel
May5-07, 12:18 PM
1. The problem statement, all variables and given/known data

The ionisation energy of Li is 5.39eV and the electron affinity of F is 3.40eV. Give a lower bound for the modulus of the crystallisation energy of Li+F-

2. Relevant equations

Not entirely sure of the equation to use, but have this one in my notes:

U = - (q1q2/r(1,2)) . alpha . (1 / 4Pi E)

where E is the permeability of free space (I believe)

Though this may be the electro-potential formula.

3. The attempt at a solution

I haven't really attempted this to any meaningful degree, the problem is that I don't know how I can work this out when I'm not given a value for r (presumably the radius between atoms or ions??) nor am I sure what values I'm supposed to use for the charges q1 and q2.

I don't expect anyone to solve this for me, but if you could clarify what you think I need to do to work this out I'd be greatful. My lecturer is great on giving derivations and formulas, but never seems to explain things very well.

Cheers
Kel

lalbatros
May5-07, 02:23 PM
What are the definitions needed to understand the question?
What is the meaning of the word affinity?
What is the ionisation energy?
What is the crystallisation energy?
Also, what is the question, why asking for a "lower bound", what do you think?
If you were asked not for a lower bound but for a good approximation of the crystallisation energy, how could you do it?

kel
May7-07, 11:18 AM
ok, from my notes there is an example using sodium chloride

which gives the energy as

E = 7.9 (cohesive energy) - 5.1(ionisation energy) +3.6 (electron affinity)
= 6.4ev

What I want to know is, is the cohesive energy generally the same for all molecules? if so, then I can use the same sort of equation to answer the question.

Otherwise, my answer would only be

E = 5.39 - 3.40
= 1.99 eV

Cheers
Kel

dimaccent
May7-07, 03:38 PM
If this is Biaggio's Assessed sheet 6, I don't think you are supposed to ask for other people to do the question for you!