Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after...

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Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are on D2L

Transcript of Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after...

Page 1: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Exam 2

Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1.

Max after the curve = 99Std Dev = 15

Grades and solutions are on D2L

Page 2: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Correspondence principleProposed by Bohr: Quantum physics results should match classical physics results in the appropriate regions (large quantum number n).

As n increases, the quantum probability averages out to flat across the well. This is exactly what is predicted by classical physics.

Only really tiny wires have noticeable quantum effects at thermal energies

Quantum probability

Classical probability

0 a

a1a2

a1a2

Quantum probability

Classical probability

0 a

n=1 n=20

Page 3: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

wire

Ene

rgy

x0 a

V(x)

Region I Region II Region IIIxDex )(III)sin()cos()(II kxBkxAx xEex )(I

)()(

II2

2II

2

xkdxxd )(

)(III

22

III2

xdx

xd )()(

I2

2I

2

xdxxd

Page 4: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

exponent must be dimensionless.

Ene

rgy

x0 a

V(x)

Particles in classically forbidden regions

Outside well: E < VInside well: E > VOutside well: E < V

Eparticle

EVm

2

What are the units of ?

A. J B. J-1 C. m2 D. m-1 E. J-1/2

xDex )(III

Page 5: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Small means slow drop off

Ene

rgy

x0 a

V(x)

Particles in classically forbidden regions

Eparticle

How far does the particle extend into the forbidden region?

A measure of the penetration depth is

EVm

2

xDex )(III

)(a

xea )(

Large means fast drop off

/1

For an electron with (V-E) ~ 3 eV this is only 10-10 m (~ size of an atom). Not very far!

Distance at which (x) is reduced by a factor of 1/e.

Page 6: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

V(x)

0 a

V(x)

0 a

EEE

Which of the four possible scenarios (A,B,C,D) would give the shortest penetration depth /1

EVm

2

xDex )(III

Eparticle

AC

BD

AD

BC

Small implies large .

Large comes from large V and/or small E

Answer C gives largest value for (V-E).

Page 7: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

wire

Consider two very long wires separated by a small gap:

wire

This is an example of a potential barrier.

Quantum tunneling through potential barrier

Quantum tunneling occurs when a particle which does not have enough energy to go over the potential barrier somehow gets to the other side of the barrier.

Eelectron

This is due to the particle being able to penetrate into the classically forbidden region. If it can penetrate far enough (the barrier is

thin enough) it can come out the other side.

Page 8: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Quantum Tunneling Sim

Page 9: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Real( )Electron can penetrate into the barrier, but is reflected eventually.

“Transmitted” means continues off to the right forever, i.e. the wave function does not go down to zero.

Electron encounters potential step

Page 10: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Real( )

Copper wire #1 Copper wire #2CuO

Quantum tunneling

If the potential increase has a finite width, it is a potential barrier and the electron can tunnel out of Region IL

L

Page 11: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Tunneling probability

The probability to tunnel across a barrier of length L is approximately:

A) P ~ L

B) P ~ -L

C) P ~ e-L

D) P ~ e-2L

decays as e-L

decays as e-2L

Page 12: Exam 2 Mean was 67 with an added curve of 5 points (total=72) to match the mean of exam 1. Max after the curve = 99 Std Dev = 15 Grades and solutions are.

Quantum tunneling probabilityThe probability of tunneling depends on two parameters:

EVm

21. The parameter measures how quickly the exponential decays and =1/ is the penetration depth (how far the wave function penetrates).

L

L

2. The width of the barrier L measures how far the particles has to travel to get to the other side.

The quantum tunneling probability is

LeP 2As increases (penetration depth decreases), probability decreases.

As L increases (barrier width increases), probability decreases.