1000 Solved Problems in Modern Physics

2.3 Solutions 113

From the geometry of the figure,

EF

CF

=

DA

CA

or

s

L+ 2 l

=

h

l/ 2

→h=

s.l

2 L+l

(8)

Eliminatinghbetween (7) and (8), we find

a=

2 sν^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,

2 s=

l(2L+l)

mν^2

μ

(

∂B

∂y

)

(11)

1 / 2 mν^2 = 2 kT

2 s=

l(2L+l)μB(∂B/∂y)

4 kT

=

[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. 2.3Stern–Gerlah
experiment

2.39 HNa
1 s 1 s^22 s^22 p^63 s