In a simple model of a potassium iodide (KI) molecule, we assume 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? ions from neutral atoms? This quantity is sometimes called the activation energy Ea. eV(b) A model potential energy function for the KI molecule is the Lennard

This question is incomplete, the complete question is;

In a simple model of a potassium iodide (KI) molecule, we assume 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- ions from neutral atoms? This quantity is sometimes called the activation energy Ea.eV

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

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

where r is the internuclear separation distance and α and ∈ are adjustable parameters (constants) . The Ea term is added to ensure the correct asymptotic behavior at large r and is activation energy calculated in a. At the equilibrium separation distance, r=r₀=0.305 nm, U(r) is a minimum, and dU/dr=0. In addition, U(r₀)=-3.37 eV.

Us the experimental values for the equilibrium sepeartion and dissociation energy of KI to determine/find 'α' and '∈'.

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

a) energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms is 1.28 eV

b) α = 0.272, ∈ = 4.65 eV

c) the force needed to break the KI molecule in nN 65.6 nN

Explanation:

a) The ionization energy of K is 4.34 ev ( energy needed to remove the outer most electrons)

And the electron affinity of I is 3.06 ev ( which is energy released when electron is added)

Now the energy that is need to transfer an electron from K to I,

i.e the ionization energy of K(4.34 ev) and the electron affinity of I (3.06 ev)

RE = 4.34 - 3.06 = 1.28 eV

b)

from the question we have

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

now taking d/drU(r₀)=0  (at r = r₀)

= 4∈d/dr [ (α/r)¹² - (α/r)⁶ ] = 0

= ( -12(α¹²/r¹³)) - (-6 (α⁶/r⁷)) = 0

12(α¹²/r¹³) = 6 (α⁶/r⁷)

α⁶ = r⁶/2

α = r/(2)^1/6

at equilibrium r = r₀ = 0.305 nm

α = 0.305 nm / (2)^1/6

C = 0.0305/1.1246

α = 0.272

Now substituting the values of U(r₀), α, Eₐ in the initial expression

U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea

we have

- 3.37eV = 4∈ [ (0.272 nm / 0.305 nm)¹² - (0.272 nm / 0.305 nm )⁶ ] + 1.28

- 1.65 eV = ∈(0.25 - 0.5)

∈ = 4.65 eV

c)

Now to break the molecule then the potential energy should be zero(0)

and we know r = 0.272 nm

therefore force needed to break the molecule is

F = -dU/dR_r-α

F = -4∈ (-12/α  + 6/α)

F = -4(4.65eV) ( -12/0.272nm + 6/0.272nm)

F = 65.6 nN

the inner core.

good luck: )

klicker