1000 Solved Problems in Modern Physics

(Tina Meador) #1

5.3 Solutions 303


5.11 Force,F=−
dV
dr


=

a
r^2


7 b
r^8
The particles will separate most easily when the force between them is a min-
imum, that is when

dF
dr

=0. This gives:
dF
dr

=−

2 a
r^3

+

56 b
r^9

= 0

r=

(

28 b
a

) 1 / 6

5.12 The inter-nuclear distance is found from


dE
dr

= 0

2 A

ro^3


8 B

ro^9

= 0 →ro^6 =

4 B

A

(1)

The dissociation energyDis formed from−D=E(ro)

−D=−

A

ro^2

+

B

ro^8

=−

A

ro^2

+

A

4 ro^2

=−

3 A

4 ro^2

where we have used (1).

A=

4 Dro^2
3
=

4

3

× 3 × 1. 6 × 10 −^19 ×(0. 4 × 10 −^9 )^2 = 1. 02 × 10 −^37

B=

A

4

r 06 =

1. 02 × 10 −^37

4

×(0. 4 × 10 −^9 )^6

= 1. 04 × 10 −^91

5.13=

(

2 kT
K

) 1 / 2

The force constantK =Ya 0 = 1. 6 × 1010 × 4. 94 × 10 −^10 = 7 .9N/m^2

T=

K

2 k

〈A〉^2 =

7. 9 ×(0. 46 × 10 −^10 )^2

2 × 1. 38 × 10 −^23

=606 K= 333 ◦C

5.3.3 Metals ...........................................


5.14 (a)vF=

(

2 EF

m

) 1 / 2

=

[

2 × 5. 52 × 1. 6 × 10 −^19

9. 11 × 10 −^31

] 1 / 2

= 1. 39 × 106 m/s

(b)τ=

m
ne^2 ρ

=

9. 11 × 10 −^31

(5. 86 × 1028 )(1. 6 × 10 −^19 )^2 (1. 62 × 10 −^8 )

= 3. 7 × 10 −^14 s
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