6.86. Suppose that Pa and Pi are the probabilities of the particle
being outside and inside the well. Then
co
2xx d x
P a I 2
Pi 1 2+33t'
a 2 sin 2 kx dx
0
where the ratio b/a can be found from the condition Vi (1) = 1)^2 (1).
Now it remains to take into account that Pa +P i = 1; then Pa =
= 2/(4 + 3n) = 14.9%.
The penetration of the particle into the region where its energy
E < U is a purely quantum phenomenon. It occurs owing to the
wave properties of the particle ruling out the simultaneous precise
magnitudes of the coordinate and the momentum, and consequently
the precise division of the total energy of the particle into the poten-
tial and the kinetic energy. The latter could be done only within
the limits set by the uncertainty principle.
6.87. Utilizing the substitution indicated, we get
x" k 2 x = 0, where k 2 = 2mE/h 2.
We shall seek the solution of this equation in the form x =
= a sin (kr + a). From the finiteness of the wave function , at
the point r= 0 it follows that a = 0. Thus, = (alr) sin kr. From
the boundary condition * (r 0 ) = 0 we obtain kr, = nn, where
n = 1, 2,... Hence, Ea = n^23 - E^2 h^2 12m,r:.
6.88. (a) * (r). I^ sin (nnrlro)
n =1, 2, ...; (b) rp,.
2nro= r (^0) /2; 50%.
6.89. (a) The solutions of the SchrOdinger equation for the func-
tion x (r):
r <r„, xi = A sin (kr cc), where k = V 2mElh,
r > ro, x2---=Bexr+ Ce-)a, where x = -1(2m (U^0 —E)1h.
Since the function AI) (r) is finite throughout the space, a = 0 and
B = 0. Thus,
A sin kr^ e r
— (^) r 2 =C r.
From the continuity of the function and its derivative at
the point r = r, we get tan kro = —klx, or
sin kro = f V h 2 12mr:U okr 0.
As it was demonstrated in the solution of Problem 6.85, this equa-
tion determines the discontinuous spectrum of energy eigenvalues.
(b) rWo = 7t 2 1i 2 /8m.
23*