Solution to 6-16
(a) For a particle in an infinite square well in its ground state, the wavefunction
is:
πx
ψ(x) = Asin( )
(1)
L
Normalization requires that:
Z
∞
2
|ψ| dx = A
1=
2
Z
−∞
Thus: A =
q
L
πx sin2
L
0
dx =
LA2
2
(2)
2
L.
The probability that the electron is within x = 0 and x = 0.1 nm is the same
as the probability that it is within x = 0 and x = L/3. So, we have:
"
√ #
Z
Z
3
2 π/3 2
2 π
2 L/3 2 πx −
= 0.1995
(3)
sin
dx =
sin θdθ =
P =
L 0
L
π 0
π 6
8
(b) Now, with n = 100 we have:
r
ψ100 (x) =
2
sin
L
100πx
L
(4)
Thus:
P =
2
L
Z
0
L/3
sin2
100πx
L
dx =
2
L
L
100π
Z
0
100π/3
sin2 θdθ =
200π
1 100π 1
− sin
= 0.3319
50π
6
4
3
(5)
(c) This holds up with the correspondence principle; as n gets large, the probability should approach 1/3, the classical probability for this situation.
1