dEldr = 0 we find reff h 2Ime2 = 53 pm, Enii,,
—me4/2h2 =
-= —13.6 eV.
6.75. The width of the image is A
6 + A'
6 -I- hl/p6,
where A' is an additional widening associated with the uncertainty
of the momentum Apy (when the hydrogen atoms pass through the
slit), p is the momentum of the incident hydrogen atoms. The function A(6) has the minimum when 6
lih//mv = 0.01 mm.
6.76. The solution of the Schrodinger equation should be sought
in the form = (x)..1 (t). The substitution of this function into
the initial equation with subsequent separation of the variables x
and t results in two equations. Their solutions are * (x) eikx,
where k =112mE1h, E is the energy of the particle, and I (t)
e -iot where co = Ern,. Finally, IF = aemix-6)0, where a is a certain constant.
6.77. P =1/3+
= 0.61.
Acos(anx11), if n =1, 3, 5,
6.78. I) = A sin (anx11), if n = 2, 4, 6,
Here A— V211.
6.80. dNIdE=(113-th)lim/2E; if E =1 eV, then dNIdE=
=0.8.107 levels per eV.
6.81. (a) In this case the SchrOdinger equation takes the form
ay
oxr a +7;_ y;....kk2t=. _ 0, k2 =2mElh2.
Let us take the origin of coordinates at one of the corners of the
well. On the sides of the well the function * (x, y) must turn into
zero (according to the condition), and therefore it is convenient to
seek this function inside the well in the form 1) (x, y) = a sin kix X
x sin k ey, since on the two sides (x = 0 and y = 0) = 0 automatically. The possible values of k1 and k2 are found from the condition of 1 turning into zero on the opposite sides of the well:
* (4, y) = 0, k1 = ± (n/h) nl, nl = 1, 2, 3, . .
(x, 12) = 0, k 2 = f (7t112) n2, n2 = 1, 2, 3, .
The substitution of the wave function into the Schrodinger equation leads to the relation lc; + k2 = k2, whence
Eno , = (n:11: + n2112 2) n2h2/2m.
(b) 9.87, 24.7, 39.5, and 49.4 units of h2/m12.
6.82. P = 1/3 — 17- 14J-c = 19.5%.
6.83. (a) E = (n2 i + n2 2
o-c2h2/2ma2, where n1, n2, n3 are
integers not equal to zero: (b) AE = n2h2Ima2; (c) for the 6-th level
74. + n2 2 n 2 2 = 14 and E = 72-c2h2/ma2; the number of states is equal
o six (it is equal to the number of permutations of a triad 1, 2, 3.)
23-9451
—me4/2h2 =
-= —13.6 eV.
6.75. The width of the image is A
6 + A'
6 -I- hl/p6,
where A' is an additional widening associated with the uncertainty
of the momentum Apy (when the hydrogen atoms pass through the
slit), p is the momentum of the incident hydrogen atoms. The function A(6) has the minimum when 6
lih//mv = 0.01 mm.
6.76. The solution of the Schrodinger equation should be sought
in the form = (x)..1 (t). The substitution of this function into
the initial equation with subsequent separation of the variables x
and t results in two equations. Their solutions are * (x) eikx,
where k =112mE1h, E is the energy of the particle, and I (t)
e -iot where co = Ern,. Finally, IF = aemix-6)0, where a is a certain constant.
6.77. P =1/3+
= 0.61.
Acos(anx11), if n =1, 3, 5,
6.78. I) = A sin (anx11), if n = 2, 4, 6,
Here A— V211.
6.80. dNIdE=(113-th)lim/2E; if E =1 eV, then dNIdE=
=0.8.107 levels per eV.
6.81. (a) In this case the SchrOdinger equation takes the form
ay
oxr a +7;_ y;....kk2t=. _ 0, k2 =2mElh2.
Let us take the origin of coordinates at one of the corners of the
well. On the sides of the well the function * (x, y) must turn into
zero (according to the condition), and therefore it is convenient to
seek this function inside the well in the form 1) (x, y) = a sin kix X
x sin k ey, since on the two sides (x = 0 and y = 0) = 0 automatically. The possible values of k1 and k2 are found from the condition of 1 turning into zero on the opposite sides of the well:
* (4, y) = 0, k1 = ± (n/h) nl, nl = 1, 2, 3, . .
(x, 12) = 0, k 2 = f (7t112) n2, n2 = 1, 2, 3, .
The substitution of the wave function into the Schrodinger equation leads to the relation lc; + k2 = k2, whence
Eno , = (n:11: + n2112 2) n2h2/2m.
(b) 9.87, 24.7, 39.5, and 49.4 units of h2/m12.
6.82. P = 1/3 — 17- 14J-c = 19.5%.
6.83. (a) E = (n2 i + n2 2
o-c2h2/2ma2, where n1, n2, n3 are
integers not equal to zero: (b) AE = n2h2Ima2; (c) for the 6-th level
74. + n2 2 n 2 2 = 14 and E = 72-c2h2/ma2; the number of states is equal
o six (it is equal to the number of permutations of a triad 1, 2, 3.)
23-9451