 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018/3/b/solution

Put

y = e^{- r / ξ}

. Differentiating the Morse potential gives

V^{'} (r) = \frac{1}{ξ} (- 2 A y^{2} + B y) .

The finite-separation minimum has

y = B / (2 A)

. Thus

a = ξ ln \frac{2 A}{B} = 0.36809 nm, V (a) = - \frac{B ^{2}}{4 A} = - 0.250 eV .

D = B^{2} / (4 A) = 0.250 eV

, measured relative to the separated-atom value

V (\infty) = 0

. This distinguishes the well's negative potential energy from the positive energy required to break the bond.

x = r - a

V (a + x) = V (a) + \frac{1}{2} κ x^{2} + O (x^{3}), κ = V^{''} (a) = \frac{B ^{2}}{2 A ξ ^{2}} = \frac{2 D}{ξ ^{2}} .

Consequently

κ = 70.862 eV n m^{- 2} = 11.353 N m^{- 1} .

The restoring force is

- κ x

to leading order, identifying the bond's Hooke's law spring constant.

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