1000 Solved Problems in Modern Physics

2.3 Solutions 119

2.52ΔE=μBBΔm=μBB(becauseΔm=±1)

=(9. 27 × 10 −^24 )(1.0)= 9. 27 × 10 −^24 J= 5. 79 × 10 −^5 eV.

The splitting of levels by equal amount in the presence of magnetic field is

called normal Zeeman effect.

f=ΔE/h= 5. 79 × 10 −^5 × 1. 6 × 10 −^19 / 6. 625 × 10 −^34 = 1. 398 × 1010 c/s

2.53 For the term^2 P 1 / 2 ,l=1,j=^12 ,s=^12 andg=^23. For the term^2 S 1 / 2 ,l=0,

j=^12 ,s=^12 andg=2. The energy levels and splitting of lines in sodium

are shown in Fig. 2.7.

2.54 The ground state energy is

E 0 =hv=

hc

λ

= 6. 63 × 10 −^34 × 3 × 1010 × 84 , 181 / 1. 6 × 10 −^19

= 10 .46 eV

The excitation linesE 2 = 10. 46 + 7. 69 = 18 .15 eV

The line 5461A is emitted when ̊ E 2 is deexcited to a lower levelE 1 such

that

E 2 −E 1 =

1 , 241

λ(nm)

=

1 , 241

546. 1

= 2 .27 eV

ThusE 1 =^18.^15 −^2.^27 =^15 .88 eV

Fig. 2.7Splitting of D 1 lines

in magnetic field

Therefore the two levels involved in the emission of the 5,461A line are ̊

18.15 eV and 15.88 eV

2.55 The 2sstate of the hydrogen atom cannot decay by electric dipole radiation

because a 2s → 1 stransition would violate theΔl =±1 rule (Laporte

rule). In point of fact the 2sstate is a metastable state with a long life time

which eventually decays to the 1sstate by a mechanism, such as collision

with other gas molecules, which is much less probable than an electric dipole

transition.