21.
(a) Using the result of problem 3 in Chapter 39,
∆E = hc
1
λ
1
−
1
λ
2
= (1240 eV
·nm)
1
588.995nm
−
1
589.592 nm
= 2.13 meV .
(b) From ∆E = 2µ
B
B (see Fig. 41-10 and Eq. 41-18), we get
B =
∆E
2µ
B
=
2.13
× 10
−3
eV
2(5.788
× 10
−5
eV/T)
= 18 T .