2.3 Solutions
113
From the geometry of the figure,
E F
C F
=
D A
C A
or
s
L +
l
2
=
h
l/2
→ h =
s.l
2L + l
(8)
Eliminating h between (7) and (8), we find
a =
2sν
2
l(2L + l)
(9)
Now the acceleration,
a =
F
m
=
μ
m
∂ B
∂ y
(10)
Finally the separation between the images on the plate,
2s =
l(2L + l)
mν 2 μ
∂ B
∂ y
(11)
1/2 mν
2
= 2kT
2s =
l(2L + l)μ B (∂ B/∂ y)
4kT
=
[0.6(2 × 1 + 0.6) × 9.27 × 10
−24
× 20]
4 × 1.38 × 10 −23 × 600
= 0.873 × 10
−2 m = 8.73 mm
Fig. 2.3 Stern–Gerlah
experiment
2.39 H Na
1s 1s
2 2s
2 2 p
6 3s
113
From the geometry of the figure,
E F
C F
=
D A
C A
or
s
L +
l
2
=
h
l/2
→ h =
s.l
2L + l
(8)
Eliminating h between (7) and (8), we find
a =
2sν
2
l(2L + l)
(9)
Now the acceleration,
a =
F
m
=
μ
m
∂ B
∂ y
(10)
Finally the separation between the images on the plate,
2s =
l(2L + l)
mν 2 μ
∂ B
∂ y
(11)
1/2 mν
2
= 2kT
2s =
l(2L + l)μ B (∂ B/∂ y)
4kT
=
[0.6(2 × 1 + 0.6) × 9.27 × 10
−24
× 20]
4 × 1.38 × 10 −23 × 600
= 0.873 × 10
−2 m = 8.73 mm
Fig. 2.3 Stern–Gerlah
experiment
2.39 H Na
1s 1s
2 2s
2 2 p
6 3s