Задачи по квантовой механике. Часть 2. Алмалиев А.Н - 12 стр.

UptoLike

w
10
(r) w
10
(r)
w
10
r
0
=
a
0
Z
f
10
(r)
R
10
(r) = r
1
f
10
(r)
R
10
(r)
1s
r
¤
n = 2
n = 2 n = 2 l = 0 m = 0
Ψ
200
n = 2 l = 1 m = 1 Ψ
211
n = 2 l = 1 m = 0 Ψ
210
n = 2 l = 1
m = 1 Ψ
211
Ψ
200
(r) = f
20
(r)Y
00
(θ, ϕ) (2s)
Ψ
211
(r) = f
21
(r)Y
11
(θ, ϕ)
Ψ
210
(r) = f
21
(r)Y
10
(θ, ϕ)
Ψ
211
(r) = f
21
(r)Y
11
(θ, ϕ)
(2p)
f
20
(r) =
1
2
µ
Z
a
0
3
/
2
exp
µ
Zr
2a
0
µ
1
Zr
2a
0
;
f
21
(r) =
1
2
6
µ
Z
a
0
3
/
2
exp
µ
Zr
2a
0
µ
Zr
a
0
.
   Òàêèì îáðàçîì, çàäà÷à çàêëþ÷àåòñÿ â ïîèñêå ìàêñèìóìà ôóíêöèè
w10 (r). Ãðàôèê ôóíêöèè w10 (r) ïðåäñòàâëåí íà ðèñ. 1.2. Ìàêñèìàëüíîå
                                                             a0
çíà÷åíèå w10 , êàê íåñëîæíî ïîêàçàòü, äîñòèãàåòñÿ ïðè r0 =      . Ýòî è
                                                             Z
åñòü íàèâåðîÿòíåéøåå óäàëåíèå ýëåêòðîíà îò ñèëîâîãî öåíòðà. Â àòîìå
âîäîðîäà îíî ðàâíÿåòñÿ áîðîâñêîìó ðàäèóñó.
   2 ñïîñîá.
   Åñëè âìåñòî ôóíêöèè f10 (r) èñïîëü-
çîâàòü R10 (r) = r−1 f10 (r), òî ôóíêöèÿ
R10 (r) áóäåò óäîâëåòâîðÿòü îäíîìåðíî-
ìó óðàâíåíèþ Øðåäèíãåðà (1.12). Èñõî-
äÿ èç âåðîÿòíîñòíîãî ñìûñëà âîëíîâîé
ôóíêöèè, ïðèõîäèì ê (1.27) è ïîëó÷àåì
òîò æå ñàìûé îòâåò.
   Çàìåòèì, ÷òî âåðîÿòíîñòü îáíàðóæå-
íèÿ ýëåêòðîíà íà ïðîèçâîëüíîì óäàëå-
íèè îò öåíòðà â 1s-ñîñòîÿíèè âñåãäà ÿâ-
ëÿåòñÿ íåíóëåâîé. Îäíàêî ïðè áîëüøèõ               Ðèñ. 1.2:
r îíà óìåíüøàåòñÿ ýêñïîíåíöèàëüíî, òàê ÷òî ðåàëüíûé ðàçìåð àòîìà áó-
äåò êîíå÷íûì.                                                         ¤

Ïðèìåð 1.5. Íàéòè ÿâíûé âèä ïîëíûõ âîëíîâûõ ôóíêöèé ñîñòîÿíèé ñî
çíà÷åíèåì ãëàâíîãî êâàíòîâîãî ÷èñëà n = 2. Ïîêàçàòü, ÷òî ýòè ôóíê-
öèè íîðìèðîâàíû íà åäèíèöó.
Ðåøåíèå. Ïðè n = 2 èìååòñÿ 4 ðàçëè÷íûõ ñîñòîÿíèÿ: n = 2, l = 0, m = 0
(Ψ200 ); n = 2, l = 1, m = 1 (Ψ211 ); n = 2, l = 1, m = 0 (Ψ210 ); n = 2, l = 1,
m = −1 (Ψ21−1 ). Íà îñíîâàíèè (1.10) èìååì:
                        Ψ200 (r) = f20 (r)Y00 (θ, ϕ)  (2s)
                        Ψ211 (r) = f21 (r)Y11 (θ, ϕ) 
                                                                         (1.28)
                        Ψ210 (r) = f21 (r)Y10 (θ, ϕ)    (2p)
                                                      
                       Ψ21−1 (r) = f21 (r)Y1−1 (θ, ϕ)
ãäå, ñîãëàñíî (1.17)
                              µ
                              ¶ 3/2    µ      ¶µ         ¶
                         1 Z              Zr         Zr
              f20 (r) = √           exp −        1−        ;
                          2a0             2a0        2a0
                          µ ¶ 3/2       µ      ¶µ ¶
                        1   Z              Zr     Zr
              f21 (r) = √            exp −             .
                       2 6 a0              2a0    a0
Îáðàòèì âíèìàíèå íà òî, ÷òî ïîñëåäíèå òðè ôóíêöèè îòëè÷àþòñÿ òîëüêî
ñôåðè÷åñêèìè ôóíêöèÿìè, íîðìèðîâàííûìè íà åäèíèöó ñîãëàñíî (1.7).


                                       12