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Energy of polarisable H2O (Read 1637 times)
Gerrit-Jan Linker
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Energy of polarisable H2O
10.02.11 at 13:02:47
 
Energy of polarisable H2O
 
The energy of H2O in which the H's are not polarisable and where O is polarisable (with polarisability α) can be constructed from the electrostatic energy and the induction energy. The H's have a partial charge of δ+ and at O there is a partial charge of δ2-.  
 
E = Uelectrostatic + Uinduction (1)
 
Uinduction is the energy that is gained when polarisation is allowed at O. It costs energy to polarise the electrons at O, to make the induced dipoles. The energy gain is that in interaction between the induced moments at O:
 
Uinduction = Upolarisation + Uinteraction (2)
 
where:
Upolarisation = +αF (3)
(This follows from the Virial Theorem)
 
Upolarisation is the energy that it costs to make induced moments at O which has a polarisability α and which is located in the field F generated by the two protons (δ+).  
 
Uinteraction = - μF. (4)
This is the general interaction energy of any dipole μ in any field F.
 
The dipole μ is the induced dipole at O due to the field generated by the H's. It is equal to the field F generated by the two H's times the polarisability of O:
μ = μinduced = αF
 
Using this in (4) we get:
Uinteraction = - μF = - αFF = -αF (5)
 
Now we can write the induction energy Uinduction (2) as:
 
Uinduction = Upolarisation + Uinteraction = +αF -αF = -αF (6)
 
What remains to complete formula (1) is the electrostatic energy.
Uelectrostatic = UH-H + 2UH-O (6)
 
Using this in formula (1) gives the final result:
 
E = Uelectrostatic + Uinduction  
  = UH-H + 2UH-O -αF      (7)
 
If Θ is the H-O-H angle we can show the angle dependency in (7) as:
 
E(Θ) = UH-H(Θ) + 2UH-O -αF(Θ)      (8)
 
See also:
Virial theorem
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« Last Edit: 23.05.11 at 21:34:43 by Gerrit-Jan Linker »  

Gerrit-Jan Linker
Linker IT Software
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