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282 8 ATOMIC STRUCTURE AND SPECTRA
E8C.12(b) Closed shells have total spin and orbital angular momenta of zero, and so do
not contribute to the overall values of S and L. (i) 3d104s2 is a closed shell
con�guration with L = 0, S = 0, and J = 0.�e term symbol is 1S0 .
(ii) For the con�guration 3d104s24p5 only the p electrons need be considered as
the others are in a closed shell. Because of the spin pairing required by the Pauli
principle the con�guration p5 has the same terms as p1: this is o�en expressed
by saying that p5 is the ‘absence’ of one electron or a ‘hole’. Holes behave just
like electrons when it comes to formulating term symbols.�erefore, one ‘hole’
in a p sub-shell has l = 1 and s = 1
2 , so L = 1, S =
1
2 , and J =
3
2 ,
1
2 . �e term
symbols are therefore 2P3/2 and 2P1/2 .
E8C.13(b) �e two terms arising from an f 1 con�guration are 2F7/2, 2F5/2, which have
S = 1
2 , L = 3 and J =
7
2 ,
5
2 .�e energy shi� due to spin-orbit coupling is given
by [8C.4–331], EL ,S , J = 1
2 hcÃ[J(J + 1) − L(L + 1) − S(S + 1)], where à is the
spin-orbit coupling constant. Hence, E3,1/2,7/2 = +(3/2)hcà , and E3,1/2,5/2 =
−2hcà .
E8C.14(b) �e selection rules for a many-electron atom are given in [8C.8–335].
(i) 2P3/2 (S = 1
2 , L = 1, J = 3
2 ) →
2S1/2 (S = 1
2 , L = 0, J = 1
2 ) has ∆S = 0,
∆L = −1, ∆J = −1 and so is allowed .
(ii) 3P0 (S = 1, L = 1, J = 0)→ 3S1 (S = 1, L = 0, J = 1) has ∆S = 0, ∆L = −1,
∆J = +1 and so is allowed .
(iii) 3D3 (S = 1, L = 2, J = 3)→ 1P1 (S = 0, L = 1, J = 1) has ∆S = −1, ∆L = −1,
∆J = −2 and so is forbidden by the S and J selection rules.
Solutions to problems
P8C.2 �e wavenumbers of the spectral lines of the H atom for the n2 → n1 transition
is given by [8A.1–304], ν̃ = R̃H(n−21 − n−22 ), where R̃H is the Rydberg constant
for Hydrogen, R̃H = 109677 cm−1. Hence, the wavelength of this transition is
λ = ν̃−1 = R̃−1H (n−21 − n−22 )−1.
�e lowest energy, and therefore the longest wavelength transition (the one
at λmax = 656.46 nm = 6.5646 × 10−5 cm) is assumed to correspond to the
transition from n1 + 1→ n1
1
λmaxR̃H
= 1
n21
− 1
(n1 + 1)2
= (n1 + 1)2 − n21
n21(n1 + 1)2
= 2n1 + 1
n21(n1 + 1)2
From the given data (λmaxR̃H)−1 = [(6.5646×10−5 cm)×(109677 cm−1)]−1 =
(7.19...)−1.�e value of n1 is found by seeking an integer value of n1 for which
n21(n1+1)2/(2n1+1) = 7.19.... For n1 = 2 the fraction on the le� is 42×92/5 =
7.2.�erefore, the series is that with n1 = 2 .