Физика твердого тела. Хуснутдинов Р.М - 8 стр.

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NaCl
a =
µz
ρN
A
1/3
.
z = 4
A
r
(Na) = 23 A
r
(Cl) = 35 M
r
(NaCl) = A
r
(Na) +
A
r
(Cl) = 58 µ(NaCl) = 58 ·10
3
a = 5.6 2 · 10
10
[231]
I = a
q
n
2
1
+ n
2
2
+ n
2
3
= 5.62 · 10
10
4 + 9 + 1 = 13.3 · 10
10
.
I = 1.33
[010] [122] [132]
[n
1
n
2
n
3
]
(hkl)
n
1
h + n
2
k + n
3
l = q,
h, k, l, q
k = q,
h + 2k 2l = q,
h + 3k + 2l = q.
h = 6 k = 4 l =
1 q = 4 {(641); 4}
x
0
= a
1
q/h =
2
3
a
1
,
y
0
= a
2
q/k = a
2
,
z
0
= a
3
q/l = 4a
3
.
     åøåíèå: Ïîñòîÿííàÿ ðåøåòêè êðèñòàëëà                      N aCl   îïðåäåëÿåòñÿ
ñîîòíîøåíèåì:
                                                  1/3
                                              µz
                                    a=                    .
                                             ρNA
Äëÿ ãðàíåöåíòðèðîâàííîé ðåøåòêè ÷èñëî èîíîâ â ýëåìåíòàðíîé ÿ÷åé-
êå z = 4. Ïîëüçóÿñü ïåðèîäè÷åñêîé òàáëèöåé Ìåíäåëååâà, íàõîäèì:
Ar (N a) = 23, Ar (Cl) = 35. Ñëåäîâàòåëüíî, Mr (N aCl) = Ar (N a) +
Ar (Cl) = 58, îòêóäà µ(N aCl) = 58 · 10−3 êã/ìîëü. Ïîäñòàâëÿÿ ÷èñ-
                                                                     −10
ëîâûå çíà÷åíèÿ â ïðåäûäóùóþ îðìóëó, ïîëó÷èì a = 5.62 · 10
ì. Ïåðèîä èäåíòè÷íîñòè êðèñòàëëà âäîëü ïðÿìîé [231]:
         q                              √
    I = a n21 + n22 + n23 = 5.62 · 10−10 4 + 9 + 1 = 13.3 · 10−10 ì.

     Îòâåò:        I = 1.33   íì.




     Ïðèìåð 3.        Íàïèñàòü èíäåêñû Ìèëëåðà äëÿ ïëîñêîñòè, ïðîõî-
äÿùåé ÷åðåç óçëû ñ èíäåêñàìè:                [010], [122], [132].    Íàéòè îòðåçêè,
îòñåêàåìûå ýòèìè ïëîñêîñòÿìè íà îñÿõ êîîðäèíàò.

     åøåíèå: Äëÿ ëþáîãî óçëà ñ èíäåêñàìè                      [n1n2 n3],   ëåæàùåãî â
äàííîé ïëîñêîñòè, èíäåêñû Ìèëëåðà                   (hkl)     óäîâëåòâîðÿþò ñîîòíî-
øåíèþ:

                                n1 h + n2 k + n3 l = q,
ãäå   h, k, l, q    öåëûå ÷èñëà. Ïîäñòàâëÿÿ â äàííîå óðàâíåíèå ïîñëåäî-
âàòåëüíî èíäåêñû âñåõ òðåõ óçëîâ, ïîëó÷àåì ñèñòåìó óðàâíåíèé:
                                
                                 k = q,
                                  h + 2k − 2l = q,
                                  h + 3k + 2l = q.
                                

åøàÿ ýòó ñèñòåìó â öåëûõ ÷èñëàõ, ïîëó÷àåì                 h = −6, k = 4, l =
−1; q = 4, ò.å. äàííàÿ ïëîñêîñòü             çàäàåòñÿ èíäåêñàìè {(641); 4}. Îíà
îòñåêàåò íà îñÿõ êîîðäèíàò îòðåçêè, ðàâíûå:

                                                2
                               x0 = a1 q/h = − a1 ,
                              
                                                 3
                                y  = a  q/k = a  ,
                               0
                                     2        2
                                z0 = a3 q/l = −4a3.

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