1000 Solved Problems in Modern Physics

114 2 Quantum Mechanics – I

Magnetic moment for both hydrogen and sodium is

1 Bohr magneton,μB=

e

2 me

= 9. 27 × 10 −^24 JT−^1

2.40 From Problem 2.38, the distance of separation on the plate

2 s=

l(2L+l)

mv^2

μB

(

∂B

∂y

)

Therefore, tanθ=L+sl/ 2 = 2 L^2 s+l=

lμB

(∂B

∂y

)

2 E =

lμB

(∂B

∂y

)

2 × 2 kT

Substitutingθ= 0. 14 ◦,l= 1 .0m,∂∂By=6Tm−^1 ,k= 1. 38 × 10 −^23 JK−^1

AndT=400 K, we findμB= 8. 99 × 10 −^24 JT−^1.

2.41 The total number of electrons is given by adding the numbers as superscripts
for each term. This number which is equal to the atomic numberZis found to
be 35. The transition elements haveZ= 21 − 30 , 39 − 48 , 72 − 80 , 104 −112,
while the rare earths comprising the Lanthanide series haveZ = 57 − 71
and actinides haveZ= 89 −100. Thus the element withZ=35 does not
correspond to either a transition element or a rare earth element.

2.42 From Fig. 2.3 of Problem 2.38 the separation of the beams as they emerge
from the magnetic field is given by
2 h=l^2 a/v^2 =(l^2 μ/mv^2 )(∂B/∂y)
=(l^2 / 4 kT)μ(∂B/∂y)
Substitutingl= 0 .07 m,μ= 9. 27 × 10 −^24 JT−^1
(∂B/∂y)=5Tmm−^1 = 5 ,000 Tm−^1 ,k= 1. 38 × 10 −^23 JK−^1 ,T= 1 ,250 K.
we find 2l= 3. 29 × 10 −^3 mor3.29 mm

2.43 (a) The magnetic moment for the silver atom is due to one unpaired electron
(b) In the^3 P 0 state the atom hasJ=0, therefore the magnetic moment is also
zero.
(c) The beam of neutral atoms with total angular momentum J is split into
2 J+1 components. 2J+ 1 =7, soJ= 3
(d) Ratio of intensities,
I 1
I 2

=(2J 1 +1)/(2J 2 +1)=

2 ×^12 + 1

2 ×^32 + 1

=

1

2

2.44 Let an electron move in a circular orbit of radiusr=^2 /me^2 around a proton.
Assume that thez-component of the angular momentum isLz=. Equating
Lzto the classical angular momentum

Lz==mer^2 ω (1)

An electron orbiting the proton with frequencyv= 2 ωπconstitutes a current

i=ω 2 πe