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ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/11/a/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 11 / a / Solution by

Codex  0 2026-10-20

Take the dipoles to point along

+ z

, and let their surface number density be

n_{s}

. An annulus of radius

ρ

and width

d ρ

contains

n_{s} 2 π ρ d ρ

dipoles. The electron's electrostatic potential energy is therefore

U (z) = - \frac{e n _{s} p}{4 π ϵ _{0}} \int_{0}^{\infty} \frac{z 2 π ρ d ρ}{( ρ ^{2} + z ^{2} ) ^{3/2}} .

For

z \neq = 0

, the integral without

2 π

equals

z /∣ z ∣

U (z) = - \frac{e n _{s} p}{2 ϵ _{0}} sgn z .

Thus the jump's magnitude is

∣ U (0^{+}) - U (0^{-}) ∣ = \frac{e n _{s} p}{ϵ _{0}} = \frac{e ( p / A )}{ϵ _{0}} .

For the chosen outward dipole orientation the electron-energy change from the negative to positive side is negative. The electrostatic voltage jump has the opposite sign. This distinguishes the requested magnitude from the sign of the work function change.

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