Coulomb attraction. Calculate the attractive (Coulomb) force between a Ca2+ ion
ID: 886372 • Letter: C
Question
Coulomb attraction. Calculate the attractive (Coulomb) force between a Ca2+ ion and an O2- ion
at a distance of 2.40 Å. If the repulsive part of the binding energy curve had the shape ER=B/rn
(n=8) and the minimum of the binding energy curve were at r0=2.40 Å, what would be the
binding energy at the minumum? (Use two infinitely separated ions as the reference (zero) of the
binding energy curve.)
Useful constants: 0=8.851012 J/(V2m), elementary charge e=1.6021019
C, 1eV = 1.6021019 J, 1Å=1010m
Hint: Write the binding energy curve in the form V(r) = A/r + B/rn, express the constant A in
units of eVÅ, and express the minimum of the binding energy curve in terms of A and r0 only
before inserting any numerical values.
Explanation / Answer
Coulomb attraction=1/(4II0 ) [ q1q2 /r20]
Calicium charge q1= +2.212C Oxygen charge q2 = - 2.212C .
= (1/4X3.14 X8.85X10-12) [+2.212CX-2.212C /(2.40X10-10m)2]
= - (1/111.156X10-12) [4.893C /5.760X10-20m]
= -4.893/111.156x5.760x10-32
= - [4.893/640.259]x1032 C2 m-2
= - 0.008x1032 C2 m-2
= -8x1029 C2 m-2
V(r) = A/r + B/rn,
-8x1029 C2 m-2=A/r + B/rn