1000 Solved Problems in Modern Physics

5.3 Solutions 301

whereEis in eV,

E=

(

0. 286

λ

) 2

=

(

0. 286

1. 66

) 2

= 0 .0297 eV.

5.7 (a)r=a/ 2

(b)r=

√

2 a/ 4

(c)r=

√

3 a/ 4

(d)

√

3 a/ 8

5.3.2 Crystal Properties ................................

5.8 Consider an infinite line of ions of alternating sign, as in Fig. 5.2. Let a nega-

tive ion be a reference ion and letabe the distance between adjacent ions. By

definition the Madelung Constant∝is given by:

Fig. 5.2Infinite line of ions
of alternating sign

α

a

=

∑

j

(±)

rj

(1)

whererjis the distance of thejth ion from the reference ion andais the

nearest neighbor distance. Thus:

α

a

= 2

[

1

a

−

1

2 a

+

1

3 a

−

1

4 a

+···

]

Or,α= 2

[

1 −

1

2

+

1

3

−

1

4

+···

]

(2)

The factor 2 occurs because there are two ions, one to the right and one to

the left, at equal distancesrj. We sum the series by the expansion:

ln(1+x)=x−

x^2

2

+

x^3

3

−

x^4

4

+··· (3)

Puttingx=1, the RHS in (3) is identified as In 2. Thus∝=2ln2.