98 Hyperfine structure and isotope shift
Fig. 6.1(a) An illustration of a nu-
cleus with spinIsurrounded by the
spherically-symmetric probability dis-
tribution of an s-electron. (b) That
part of the s-electron distribution in the
regionr<rbcorresponds to a sphere
of magnetisationManti-parallel to the
spins.
(b)(a)of the total. For s-electrons this distribution is spherically symmetric
and surrounds the nucleus, as illustrated in Fig. 6.1. To calculate the(^2) See Blundell (2001) or electromag- field atr= 0 we shall use the result from classical electromagnetism 2
netism texts. that inside auniformlymagnetised sphere the magnetic flux density is
Be=
2
3
μ 0 M. (6.5)However, we must be careful when applying this result since the distri-
bution in eqn 6.4 is not uniform—it is a function ofr. We consider the
spherical distribution in two parts.
(a) A sphere of radiusr=rb,whererba 0 so that the electronic
wavefunction squared has a constant value of|ψ(0)|^2 throughout
this inner region, as indicated on Fig. 6.2.^3 From eqn 6.5 the field(^3) This spherical boundary atr =rb
does not correspond to anything physi-
cal in the atom but is chosen for math-
ematical convenience. The radiusrb
should be greater than the radius of the
nucleusrNand it easy to fulfil the con-
ditionsrNrba 0 since typical nu-
clei have a size of a few fermi (10−^15 m),
which is five orders of magnitude less
than atomic radii.
inside this uniformly magnetised sphere is
Be=−
2
3
μ 0 gsμB|ψns(0)|^2 s. (6.6)(b) The part of the distribution outside the spherer>rbproduces no
field atr= 0, as shown by the following argument. Equation 6.5
for the field inside a sphere does not depend on the radius of that
sphere—it gives the same field for a sphere of radiusrand a sphere of
radiusr+dr. Therefore the contribution from each shell of thickness
dris zero. The regionr>rbcan be considered as being made up of
many such shells that give no additional contribution to the field.^4(^4) The magnetisation is a function of
ronly and therefore each shell has a
uniform magnetisationM(r) betweenr
andr+dr. The proof that these shells
do not produce a magnetic flux den-
sity atr= 0 does not requireM(r)
to be the same for all the shells, and
clearly this is not the case. Alterna-
tively, this result can be obtained by in-
tegrating the contributions to the field
at the origin from the magnetic mo-
mentsM(r)d^3 rover all angles (θand
φ).
Putting this field andμIfrom eqn 6.1 into eqn 6.3 gives
HHFS=gIμNI·
2
3
μ 0 gsμB|ψns(0)|^2 s=AI·s. (6.7)