3)
The wave function for a particle confined to moving in a 1D box is given by:
A = Ö2/ ÖL
(Refer to the normalization condition applied to the system in 5(b), i.e. solving for A)
4) From the information and the Heisenberg Uncertainty Principle:
DP x (D x) > h/ 2p = 1.05 x 10-34 J-s = 1.05 x 10-27 erg-s (cgs units)
The uncertainty in the x-component is 0.5 angstrom where 1 A = 1.0 x 10-8 cm
Then the uncertainty in the x-component is: Dx = 0.5A = 5.0 x 10-9 cm
The uncertainty in the x-component of the momentum of the electron is:
DP x» (h/ 2p) / D x = 1.05 x 10-27 erg-s/ 5.0 x 10-9 cm
Or, 2 x 10-19 g cm-s
5 (a)) In classical theory the probability that the particle would be in the region between x and x + L/3 is 0.33. Since no other information is given then it is equally probable that it’s in any place in the region – so one third of the time will be spent over any given line segment, i.e. which is 0ne-third the total length.
b) For a quantum mechanical interpretation, define the wave function (See e.g. http://brane-space.blogspot.com/2013/05/solutions-to-quantum-box-problems.html:
y = A sin 2kx
sin 2kL = 0 and k (wave number ) = n p/L
y = A sin (n px/L)
For normalization:
1 = = ò L 0 P(x) dx = A2 ò L 0 sin 2 (n px/L) dx
A2 (L/ n p) [ u/2 – sin 2u/ 4] 0 n p = 1
Then the probability it will lie between 0 and L/3 is:
ò L/3 0 sin2 (npx/L) dx
= 2/ n p [n p/ 6 - ¼ (sin 2 p/ 3 )] =
2/ p [p/ 6 - ¼ (sin 2 p/ 3 )] = 0.19
2/ p [p/ 6 - ¼ (sin 2 p/ 3 )] = 0.19
(Since n = 1 is the lowest energy state)
c) Since n = 2 is the second lowest state the probability the particle is between 0 and L/3 is:
(2/L) L/ 2 p [p/ 3 - ¼ (sin 4 p/ 3 )] = 0.40
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