MSE 410/ECE 340 Electrical Properties of Materials …coen.boisestate.edu/bknowlton/files/2011/09/MSE-Problem...MSE 410/ECE 340 School of Materials Science & Engineering Electrical

MSE 410/ECE 340 School of Materials Science & Engineering Electrical Properties of Materials Fall 2017/Bill Knowlton

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Problem Set 5 Solutions (Show all work for all problems)

1. The Heisenberg uncertainty principle is given by:

2xx p

where x and px are the standard deviation of position and momentum, respectively. They are determined via their expectation values of x and px. Specifically, they are given by:

22

22

ˆ ˆ

and

ˆ ˆx x x

x x x

p p p

For the particle in the box problem, determine xp by finding each expectation values separately.

Simplify your answers as much as possible. The wave function for the particle in the box problem

can be written as: ( )n

x AiSin xa

. You will need to normalize it; however, if you have

already normalized it in a previous problem set, then you do not need to normalize it again. Just use the answer you had obtained previously.

Note: This is a math intensive problem, so do not take it lightly.


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2. In class, we did the particle in the box problem. For the particle in the box problem, do the following:

a. Plot the wave functions as a function of x in a 1 nm wide box for n = 1 - 4. (Hint: see Kasap, fig. 3.15)

In both part a and part b, a = 1 nm.

Figure: The wave function for a particle in a box of width 1nm for n = 1-4. The wave function is pinned (i.e., has a node) at x = 0nm and x=1 nm which forms a standing wave. Each wave has n+1 nodes including the nodes at each end.

b. Plot the probability densities as a function of x in a 1 nm wide box for n = 1 - 4. (Hint: see

Kasap, fig. 3.15)

Figure: The magnitude squared, or probability density, of a wave function for a particle in a box of width 1nm for n = 1-4. The magnitude squared of the wave function is pinned (i.e., has a node) at x = 0nm and x=1 nm which forms a standing

0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

14

x nm

n1

,n

2,

n3

,n

4

x sin n x a ; Note : a 1 nm

0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

14

x nm

2n

1,

2n

2,

2n

3,

2n

4

Probability Density x 2; Note : a 1 nm


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wave. While the nodes of the magnitude squared of each wave has n+1 nodes including the nodes at each end, the magnitude of the wave is always positive since the magnitude is squared. In other words, the probability of finding an electron cannot be negative.

c. Comment on your findings in part a and b.

In part a, we see the following:

Standing waves are apparent

The wave functions all go to zero at the boundaries as they should.

At the nodes, the displacement of the waves are zero

The ground state wave function does not have nodes

Wave functions for which n is odd are symmetric about a/2

Wave functions for which n is even are asymmetric about a/2

In part b, we see the following for the probability density, p(x). Note that:

p(x) = ψ *ψ

All P(x)'s go to zero at the boundaries as they should.

The probability density at the nodes is 0. That is, the probability of finding an electron at the position of the nodes is zero.

If one adds the area under each curve for each p(x), it should equal one if the wave functions are normalized.

The magnitude is always positive since the probability of finding an electron should never be negative.


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3. Solve the infinite (in positive x direction) potential barrier step problem for Ee- < Vo. Solve this problem thoroughly without skipping any steps using the approach we listed and outlined in class including: Draw the schematic diagram of the potential barrier that includes labeling V, x, regions,

boundaries, wave functions for each region and associated coefficients and arrows, and special wave functions.

The case (here, it is Ee- < Vo) For each region, write down the wave functions and Hamiltonians and solve the S.E. to

determine the eigenenergies and then the k’s. Use or apply the boundary conditions (B.C.s) to solve for several of the coefficients or their

ratios Calculate T and R step by step using J’s When calculating T and R step by step using J’s, do not use the relationship T + R = 1 to solve the problem, although you can use it to check your answers. Solutions


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4. For the infinite (in positive x direction) potential barrier step problem for the case Ee- > Vo, do the following: a. Draw the schematic diagram of the potential barrier that includes labeling V, x, regions,

boundaries, wave functions for each region and associated coefficients and arrows, and special wave functions.

b. For each region, directly write down the wave functions, energies (eigenenergies), and k’s. In

other words, you do not have to solve the S.E. or apply boundary conditions.

c. Calculate JI, JR and JT showing all steps. You may have calculated J, for some wave function,

in a previous problem set.


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d. Using part c, calculate T and R.

e. Plot T and R as a function of the electron energy Ee- in units of eV. Use a value of 1 eV for

the potential barrier. Recall that this is for the case Ee- > Vo. T and R as a function of the electron energy Ee- are given by:

2

2 2

14;

1 1

: 1 o

e

T R

Vwhere

E

Figure Caption: Both T(E) and R(E) are plotted as a function of electron energy for the case in which Ee- > Vo where the potential energy step, Vo, is 1eV. Note that the plots do not show classical behavior of a step function at E=Vo=1eV to T(E) = 1 or R(E)=0.


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f. Thoroughly comment on the outcome and findings of the plots.

It can be observed from the plot that as the electron energy increases, T(E) approaches 1 and R(E) approaches 0. Additionally, as the electron energy decreases, T(E) approaches 0 and R(E) approaches 1. From the viewpoint of classical physics, T(E) should always be 1 and R(E) should always be 0. In the electron energy range of approximately between 1-5 eV, we see that the classical viewpoint is not met and this is due quantum mechanical effects – i.e., the wave-like nature of electron motion. For electron energies beyond 5 eV, T(E) approaches 1 which is essentially a classical behavior and quantum effects are essentially not observed. And for both descriptions, the relationship T + R = 1 holds.


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5. Consider a finite electron energy barrier of energy, Vo, that has a thickness, ao. The problem is similar to the problem in your text book by Kasap. Consider the case that Ee- > Vo, where Vo is 3.1eV.

1 1 1 1

2 2

3 1 1

2 21

1

2 22

2

2 21

3

Region 1: ( ) ; & ( ) & ( )2

Region 2: ( ) ; 2

Region 3: ( ) ; & ( )2

ik x ik x ik x ik xI R

ik x ik x

ik x ik x ik xT

kx Ae Be E x Ae x Be

m

kx Ce De E V

m

kx Fe Fe E x Fe

m

T(Ee-,a) is given by:

2 22

1 2

24 2 2 4 2 2

1 1 2 2 2 1 2

8FT

A [ ] 6[ ] [ ] [ ] [4 ] [ ] [ ]

e e

e e e e e e e

k E k E

k E k E k E k E Cos ak E k E k E

Note that k1 and k2 are a function of e

E [i.e., 1 1 2 2 and e e

k k E k k E ].

For this problem, plot the transmission coefficient, T(Ee-,a), as a function of electron energy, Ee-, in eV for the following values of the barrier thickness, a: 0.5 nm and 5 nm. Again, the potential barrier energy, Vo, is 3.1eV. Include both plots on one graph where the electron energy axis is from 0 to 10 eV (however, keep in mind that T(Ee-,a) is for the case of Ee- > Vo). Include a figure caption. Please provide thoughtful insight and comments concerning your observations and findings of the behavior of T(E) in the plot.

Solution:


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Figure Caption: A plot of the transmission coefficient, T(Ee-,a), as a function of electron energy, Ee-, in eV for the following values of the barrier thickness, a, 0.5 nm and 5 nm, where the potential barrier energy, Vo, is 3.1eV.

Comment on the observations and findings of the plot: Overall: Although Ee- > Vo, T is only 1 for specific values, otherwise T < 1. This is due to the fact the (x) is a plane wave which is a sinusoidal function. The potential, Vo, that is buried in k2 acts to damp the plane wave. Physically, what this is telling us is that the wave is being diffracted.

For changes in a: As a increases in thickness, the oscillatory behavior of the transmission coefficient, T, and the amplitude of T increase. Thus, as a increases, there are certain energies at which probability for electron transmission is lower than for thinner a’s, which is due to the increase in amplitude. The increase in the oscillatory behavior as a increases results from the increase in interaction of the electron with the potential. All effects are due to the quantum mechanical effect of the wave-like nature of the electron.

The oscillatory behavior is the electron being diffracted by the potential barrier. Note at higher electron energies, the attraction by the potential barrier is significantly reduced and approaches the classical picture where diffraction does not occur.

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MSE 410/ECE 340 Electrical Properties of Materials …coen.boisestate.edu/bknowlton/files/2011/09/MSE-Problem...MSE 410/ECE 340 School of Materials Science & Engineering Electrical