3)
The wave function for a particle confined to moving in a 1D box is given by:

**: A =**

**Ö**

**2/**

**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)

^{ 2 }= A

^{2}sin

^{2}(n px/L)

For normalization:

1 = =

**ò**^{L }_{0}_{ }P(x)^{ }dx = A^{2}**ò**^{L }_{0}_{ }sin^{2}(n px/L) dx
A

^{2}(L/ n p) [ u/2 – sin 2u/ 4]_{ 0}^{n }^{p}^{ }^{ }= 1^{2}(L/ n p) (n p/2) = 1

^{2}= 2/L so that:

**A =**

**Ö**

**2/**

**ÖL**

Then the probability it will lie between 0 and L/3 is:

**ò**

^{L/3 }

_{0}_{ }sin

^{2}(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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