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irodov problems in atomic and nuclear physics

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-- n~n2 ) ;
21(b) (T) == T 1i2/mr~; (e) expand the function ~'1s (r) in terms
of eigenfunctions of the operator k:
~)ts (r) = (2:rr) -3/2 i Cke i k r dk,
where Ck = (2:rr)-3/2 ) '¢1s (r) e- i k r dV. The latter integral is to be
computed in spherical coordinates with the polar axis being
directed along the vector k:
1 eJ\ sin (nr/ro> ikr cos t} 2 d . ~<\, d~<\' dCk == , / \ e- r r sm tr ·v <p(2n)2 J" ro r
v; sin kro
k (n 2 - k2r8) •
Therefore the probability ~density of the given wave vector is:
ICkl2 = k 2 [;2s~;:'i)2 . To find the probability of (the modulus
of the wave vector being between the values k and k +dk,
integrate the latter expression with respect to all possible-
directions of the vector k, i ,e, mi1tiply it by 4nk2 dk: Finally
we get
w (k) dk = 4nro sin2 kro dk.
(n 2- k2r8)2
4.68. (a) The solutions of the Schr6dinger equation for the-
function X (r) are
r<ro, Xt=Asin(kr+a), k=l/2inE/1i;
r ;» ro, 'X2 = Bexr +Ce:>: x = V2m (Vo- Evlh,
From the finiteness of the function '¢ (r) throughout the space,
it follows that a = 0 and B = O. Thus,
Ilh -A sinkr. Ilh -C e- x r
~1- -r-' ~2- -r-·
From the condition of continuity of 'tjJ and 'P' at the point r = ro,·
we obtain tan kro = -k/x, or sin kro = +V 1i2/2mr~Uokro. This
equation, as it is shown in the solution of Problem 3.47, defines the
discrete-spectrum of energy eigenvalues.
(b) n2Ji2/8m < r~Uo< 9n2Ji2/8m. (c) In this case there is a single
3 V3 2 211 2 1i2level: sinkro=-4-kro; kr o= -3 n: E===-o-·--2. From theJt ~J mro
condition a(r2'¢2)18r == 0, we find r p r = 3ro/4; 34%.
.4.69. iJiJ2~ +~ aaR ·t(E+~- l(lt 1)) R=O, p=rlrt, E=
p P P P P
~E/Ei·
4.70. (a) Neglecting the small values, reduce the Schrodinger
equation to the form X" - X2X = 0, X = V 2m IE I/h. Its solution
is X (r) = Aexr + Berw. From the finiteness of R (r), it follows that
A = 0 and R (r) 0: e-xr/r. (b) Transform the Schrodinger equation
to the form X" - l (l t 1) X = O. I ts solution is to be found in the
r
form X = Ara . After the substitution into the equation, we find
-two values of a: 1 + land -l. The function R (r) is finite only if
o: = 1 + l. Hence, R (r) 0: r l •
4.71. (a) Substituting this function into the Schr6dinger equa-
tion, we obtain B (a, ex, E) + rC (a, ex, E) + ~D (a, ex) = 0, where
r
B, C, and D are certain polynomials. This equality holds for any
values of r only when B = C = D = 0, whence
a = a = -1/2r1 = -me2/21i2 ; E = -me4/81i2•
(b) A = -} (2nr~) -1/2; rt is the first Bohr radius.
4.72. (a) rp r = r1 , where r1 is the first Bohr radius; 32.3%;(b) 23.8%.
4.73. (a) (r) == 3rl/2; (r2 ) == 3r~; «L\r)2) == (r2) - (r)2 == 3r~/4; r. is
the first Bohr radius; (b) (F) ~ 2e2/r~; (U) ~ -e2/rt; (c) (T) =
= J1~f'¢ dT = me"12/i2; V(v2) = e2//i = 2.2.106 m/s.
4.74. (a) 4rt and 9r1; (b) 5r~ and 15.75r;; r1 is the first Bohr
.radius.
4.75. GOo "CO' ) P~r) 4nr2dr = :1' where p (r) = e'¢rs (r) is the
-vclume density of charge; rt is the first Bohr radius.
4.76. Write Poisson's equation in spherical coordinates:
i.. a0 22 (rcpe) = 4~ecpr8 (r), e ;» O.
r r
Integrating this equation twice, we get
q>e (r) = ( :1 + ; ) e- 2r/T1 + .4+ ~ ,
where r 1 is the first Bohr radius. A and B are the integration con-
.stants. Choose these constants so that cP e (00) = 0 and cP e (0) be finite.
Hence, A = 0, B = -e. Adding the potential induced by the nuc-
·176
j
5/2
3/2
1/2
3/2
Fig. 66
J
I 2 J 4 5
~
=2
n=
Ieus to the expression obtained, we get
cp (r) ~ ( ..:.. +~ ) e - 2 r [r 1.
, rl r
4.77. See the solution of Problem 4.67, (c), w(k)dk===
32rtk2 dk . .
= 1t (1+k2r;)-l , where r 1 IS the first Bohr radi us.
5.1. 5.14 and 2.1 V.
5.2. 0.41, 0.04, and 0.00.
5.3. Having calculated the quantum defect of S terms, we find
E b = 5.4 eVe
5.4. (a) 6; (h) 12.
5.5. O.2i and 0.05; 0.178 urn.
5.6. a == 1.74; n = 2.
5.7. 7.2.10-3 eV; 1.62 -v.
5.8. 555 cm -1.
5.tO. (a) ~T === a 2RZ4 (n - 1)/n4 === 5.85. 2.31 and 1.10 cm r-;
(b) 1.73 and 0.58 em -I (three sublevels).
5.11. ~A. == a 2/9R = 5.4.10-3 A (equal
for Hand He+).
5.12. Z === 3, i.e. Li"".
5.13. (a) See Fig,: 66; dVS1 = vs-v t ===
=7.58 em-I, L\A.s1 = 0.204 A; (b) ~v ===
= 2.46 em-I, dA = 0.54 A.~
5.14. A/fJA. > v/(v3 - v2) === 4.2.105(see Fig. 66).
5.15. In units of n. V3tj/2, V15/2, n
and 1/3/2 (4.P); 2 V5, 2 V3, VEi, V2,
and 0 (5D).
5.16. (a) 1Pt and 3Po, t, 2; (b) ipf,
1D2 , 1F3, 3Po. 1,2' 3D1, 2. 3' 3F2 , 3, 4.; (c) 2P1/ 2 , 3/ 2 , 2D3 / 2 , 5 / 2 ,
2F5/ 2 , 7/ 2 , 4.P1/ 2 , 3/2. 5/2, 4.D1/2 , 3/2,5/2,7/2, 4F3/ 2 , 5/2,7/2, f)/2·
.5.17. 20 (5 singlet and 15 triplet types).
5.18. ISo, IP1 , ID 2 , 3S1 , 3PO, l , 2 ' 3D1,2t3.
5.19. (a) 2,4, 6,8; (b) respectively, 2; 1, 3; 2, 4; 1, 3, 5.
5.20. V301i.
5.21. Respectively, Ps>V2nand Ps:::= V2/i.
5.22. (a) 35.2°; (b) 34.4°.
5.23. 10 (the number of states with different values of mJ).
5.24. Jtr 30 n ; 5113 ,
5.25. 125°15'.
5.26. (a) ~ (2/ + 1) = (28 + 1)· (2£ + 1); (b) 2 (2l 1 + 1) X
J
X 2 (2l 2 + 1) = 60; (c) the number of states with identical quan-
12-0339 177
tum numbers nand l is N = 2 (2l + 1). While distributing k
electrons over these states, the Pauli exclusion principle should he
taken into account. Consequently, the problem reduces to finding
the number of combinations of N elements taken k at a time:
Ck _ N(N-1)(N-2) ... (N-k+1) -120N- kf -.
5.27. (a) 15; (b) 46.
5.28. (a) 2 (2l + 1); (b) 2n2•
5.29. (a) C: 1s22s22p2(3Po); N: 1s22s22p3(4S3/2);
(b) S: 1s22s22p63s23p4(3P2); CI: 1S22s22pG3s23p5(2P3/2).
5.30. (a) 3F2; (b) 4F3/2.
5.31. 685/ 2 •
5.32. The basic term 5D". The degeneracy 2J + 1 = 9.
5.33. Let us compile the table of possible distributions of elec-
trons over quantum states (numbers) with the Pauli exclusion
principle taken into account (Tables 1 and 2). While doing this,
we can leave out those distributions that provide the negative values
of the sum of projections M Land M s; such distributions do not
submit anything new, which can be proved directly.
To illustrate, let us denote the spin projection m ; of each electron
by the arrow pointing either up (if m , = +1/2) or down (if
m , = -1/2).
Table 1
+1 t t t t t.J. -
0 t - .J. - - tt
-1 - t - t - -
Ms 1 1 0 0 0 0
M L 1 0 1 0 2 0
Table 2
+1 t t t t t+ t+ t
0 t t t t t - t+
-1 t .J. t t - t -
Ms 3/2 1/2 1/2 1/2 1/2 1/2 1/2
M L 0 0 0 0 2 1 t
(a) See Table 1. The presence of the state with M L === 2 and
Al s == 0 indicates that there is a term ID; consequently, there must
also be two other states: M L == 1 and M L == 0 (for both ill s == 0).
From other distributions the state with M L = 1 and M s = 1
points to the existence of the 3p term; therefore, there must be still
another state with M L = 0 and M s = 1. The last remaining state
with M L = 0 and M s = 0 belongs to the IS term. Consequently,
three types of terms correspond to the given configuration: 18, ID,
and 3P.
(h) See Table 2. Via similar reasoning, we get 2D, 2p, and 48.
(e) 18, ID, IG, 3p, and 3F.
178
5.34. Both configurations have the following identical type!
of terms: (a) 2p; (b) IS, ID, and 'JP; (c) 2D. This follows from the fact
that the absence of an electron in a snbshell can be treated as "a hole"
whose state is determined by the same quantum numbers as those
of the absent electron.
5.35. Let us compile the table of possible distributions of elec-
trons over quantum states, taking into account that the Pauli
exclusion principle imposes limitations only on equivalent elec-
trons.
(a) See Table 3 in which the thin arrows indicate spin projections
of a p-electron and heavy arrows those of an s-electron.
Table 3
+1 t t t t ... t t+ -I
0 tt tt t -!- it t t t-!-t
-1 - - t J.. - -!- - -
ills 3/2 If' 1/2 3/2 1 .) 1 ") 1/2 1/2 1/2, .... / ....
~,,, L 1 1 0 () 1 0 2 0
t'
The possible types of terms: 2D, 2p, 28, and 4P.
(b) 28,