In the potassium iodide (KI) molecule, vassume the K and I atoms bond ionically

Question

         In the potassium iodide (KI) molecule, vassume the K and I atoms bond ionically by the transfer of one electron from K to I.

(a) The ionization energy of K is 4.34 eV, and the electron affinity of I is 3.06 eV. What energy is needed to transfer an electron from K to I, to form K+ and I- from neutral atoms (in eV) . This quantity is sometimes called the activation energy Ea.

(b) A model potential energy function for the KI molecule is the Lennard -Jone potential:

(U(r)=4epsilonleft[    left( rac{sigma}{r} ight)^{12}  - left( rac{sigma}{r} ight)^{6}   ight]+E_a)

where r is the internuclear separation distance and epsilon and sigma are adjustable parameters. The Ea term is added to ensure the correct asymptotic behavior at large r. At the equilibrium separation distance, r=ro=0.305 nm, U(r) is a minimum, and dU/dr=0. In addition, U(ro)=-3.37 eV. Find sigma and epsilon.

signma in nm

epsilon in eV

(c) calculate the force needed to break the KI molecule in nN

(d) Calculate the force constant for small oscillations about r=ro. Suggestion: Set r=ro+s, where S/ro<<1, and expand U(r) in powers of s/ro up to the second order terms. The answer will be in N/m

Please show all working - including formula used

Explanation / Answer

part c)

F = -dU/dr = 4 epsilon (12 * sigma^12 r^(-13) - 6 * sigma^6 r^(-7))

we want to find F_max:

==> dF/dr = 0

==> 4 epsilon (-156 * sigma^12 r^(-14) + 42 * sigma^6 r^(-8)) = 0

==> r = (156/42)^(1/6) sigma = 3.381e-10 m

==> F(r) = 6.55 nN

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