Lecture 37 - University of Coloradojcumalat/phys2170_f13/lectures/Lec37.pdfwith the quantum numbers...

http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 1

Lecture 37

•  Will do hydrogen atom today •  After Thanksgiving break we have only two weeks

before finals. We will talk about multielectron atoms, Pauli Exclusion Principle, etc. – up thru Chapter 10.

•  A few interesting facts for Friday!

http://www.colorado.edu/physics/phys2170/

Renewable Source of Energy – Fun Facts

Physics 2170 – Fall 2013 2

Club Watt in Rotterdam, Netherlands is using floor vibrations from people walking and dancing to power its light show. The vibrations are captured by “piezoelectric” materials that produce an electric change when put under stress.

The U.S. Army is also looking at piezoelectric technology for energy. They put the material in soldier’s boots in order to charge radios and other portable devices. Although this is an interesting renewable energy with great potential, it’s not cheap.


Four Motion Sensor Types


1) Passive infrared sensors detect a person's body heat as it changes against the background of the room. No energy is emitted from the sensor, thus the name "passive infrared). Humans, having a skin temperature of about 93 degrees F, and radiate with a wavelength between 9 and 10 micrometers. Therefore, the sensors are typically sensitive in the range of 8 to 12 micrometers. 2) Sends pulses of ultrasonic waves (above the frequency that a human can hear) and measures the reflection off a moving object. Motion causes the frequency of the reflected wave to change (Doppler effect). 3) A microwave sensor sends out electromagnetic pulses and measures the changes in frequency (Doppler) due to reflection off a moving object. 4) Tomographic motion detection systems sense disturbances to radio waves as they pass from node to node of a mesh network. They have the ability to detect over complete areas because they can sense through walls and obstructions.


Hot and Cold Running Water - Infrared



Shoe after just being worn -infrared



Tomographic Detection


Detects the presence of humans based on changes in the baseline signal strength between nodes. The advantage is the ability to pass through walls, furniture and other obstructions. This is implemented using signals in the 2.4 GHz range


Thanksgiving Facts


The pilgrims didn't use forks; they ate with spoons, knives, and their fingers.

The first Thanksgiving celebration lasted three days.

Benjamin Franklin wanted the turkey to be the national bird of the United States.

Turkeys have heart attacks. The United States Air Force was doing test runs and breaking the sound barrier. Nearby turkeys dropped dead with heart attacks from the shock wave.

Abraham Lincoln issued a 'Thanksgiving Proclamation' on third October 1863 and officially set aside the last Thursday of November as the national day for Thanksgiving.

Sarah Josepha Hale, an American editor, persuaded Abraham Lincoln to declare Thanksgiving a national holiday. She is author of the nursery rhyme "Mary Had a Little Lamb"


3-D central force problems The hydrogen atom is an example of a 3D central force problem. The potential energy depends only on the distance from a point (spherically symmetric)

x y

z

θ

φ

r

Spherical coordinates is the natural coordinate system for this problem.

General potential: V(r,θ,φ) Central force potential: V(r)

€

−2

2µ1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sinθ

∂∂θ

sinθ ∂ψ∂θ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin2θ

∂2ψ∂φ 2

⎡

⎣ ⎢

⎤

⎦ ⎥ +V (r)ψ = Eψ

The Time Independent Schrödinger Equation (TISE) becomes:

We can use separation of variables so

CM of mass motion is ignored!


Separation of Variables


€

−2

2µ1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sinθ

∂∂θ

sinθ ∂ψ∂θ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin2θ

∂2ψ∂φ 2

⎡

⎣ ⎢

⎤

⎦ ⎥ +V (r)ψ = Eψ

€

−sin2θ ∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ − E −V (r)[ ]ψ 2µr

2 sin2θ2

− sinθ ∂∂θ

sinθ ∂ψ∂θ

⎛

⎝ ⎜

⎞

⎠ ⎟ =

∂ 2ψ∂φ 2

Divide thru by Ψ=RθΦ and set both sides equal to a constant, -m2, Separates r and theta side from phi side.

€

−sin2θR

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ − E −V (r)[ ] 2µr

2 sin2θ2

−sinθΘ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ = −m2 =

d2Φdφ 2


Angular momentum about the z-axis is quantized:

Note is similar to which is the solution to the free particle with

Angular momentum quantization about z-axis

x y

z

θ

φ

r As k gives the momentum in the x direction, m gives the momentum in the φ direction (angular momentum).

There is nothing truly special about the z-axis.

We can point the z-axis anywhere we want to. It is just the nature of the coordinate system that treats the z-axis differently than the x and y axes.


The Θ(θ) part

x y

z

θ

φ

r

The solution to the Θ(θ) part is more complicated so we skip it.

The end result is that there is another quantum variable ℓ which must be a non-negative integer and ℓ ≥ |m|.

The ℓ variable quantizes the total angular momentum:

Note, for large ℓ, so ℓ is basically the total angular momentum and m is the z-component of the angular momentum.

Since the z-component cannot be larger than the total, |m| ≤ ℓ.

Solving for Θ, and requiring behavior Is finite everywhere forces

€

α = ( +1)

€

= m ,m +1,m + 2,m + 3.....where


Angular Momentum Picture



Spherical harmonics We have determined the angular part of the wave function so

has become with the quantum numbers ℓ and m specifying the total angular momentum and the z-component of angular momentum.

This angular solution works for any central force problem.

The combination are the spherical harmonics


The radial component of ψ For any central force potential we can write the wave function as

To solve this equation we need to know the potential V(r).

For the hydrogen atom

The radial part of the time independent Schrödinger equation can be written as

This is how we are going to get the energy E and the r dependence of the wave function

Note that m does not appear. This makes sense because it just contains information on the direction of the angular momentum.

The total angular momentum is relevant so ℓ shows up.


Clicker question 1 Set frequency to DA For any central force potential we can write the wave function as

Q. What are the boundary conditions on the radial part R(r)? A. R(r) must go to zero as r goes to 0 B. R(r) must go to zero as r goes to infinity C. R(∞) must equal R(0) D. R(r) must equal R(r+2π). E. More than one of the above.


Clicker question 1 Set frequency to DA For any central force potential we can write the wave function as

Q. What are the boundary conditions on the radial part R(r)? A. R(r) must go to zero as r goes to 0 B. R(r) must go to zero as r goes to infinity C. R(∞) must equal R(0) D. R(r) must equal R(r+2π). E. More than one of the above.

In order for ψ(r,θ,φ) to be normalizable, it must go to zero as r goes to infinity. Therefore, R(r)→0 as r→∞.

Physically makes sense as well. Probability of finding the electron very far away from the proton is very small.


Radial Equation


€

1r2

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +2µ2

E −V (r)[ ]R = ( +1) Rr2

In order to prevent this equation from diverging as r->∞, it is found that an integer n must have a value and

€

n = +1, + 2, + 3,...

€

En = −µZ 2e4

22n2

Schrodinger’s derivation of this equation in 1926 constituted the first convincing evidence of quantum mechanics

Wavefunction depends on all three quantum numbers, but Energy only depends on the quantum number n.


The three quantum numbers For hydrogenic atoms (one electron), energy levels only depend on n and we find the same formula as Bohr:

For multielectron atoms the energy also depends on ℓ.

There is a shorthand for giving the n and ℓ values.

n = 2 ℓ = 1

Different letters correspond to different values of ℓ s p d f g h … 0 1 2 3 4 5


Hydrogen ground state The hydrogen ground state has a principal quantum number n = 1

Since ℓ<n, this means that ℓ=0 and therefore the ground state has no angular momentum.

Since |m|≤ℓ, this means that m=0 and so the ground state has no z-component of angular momentum (makes sense since it has no angular momentum at all).

Note that Bohr predicted the ground state to have angular momentum of ħ which is wrong. Experiments have found that the ground state has angular momentum 0 which is what quantum mechanics predicts.


Clicker question 2 Set frequency to AD

n = 1, 2, 3, … = Principal Quantum Number

ℓ = 0, 1, 2, … n-1 = angular momentum quantum number = s, p, d, f, …

m = 0, ±1, ±2, … ±ℓ is the z-component of angular momentum

A hydrogen atom electron is excited to an energy of −13.6/4 eV. How many different quantum states could the electron be in? That is, how many wave functions ψnℓm have this energy?

A.  1 B.  2 C.  3 D.  4 E.  more than 4


Clicker question 2 Set frequency to AD

n = 1, 2, 3, … = Principal Quantum Number

ℓ = 0, 1, 2, … n-1 = angular momentum quantum number = s, p, d, f, …

m = 0, ±1, ±2, … ±ℓ is the z-component of angular momentum

A hydrogen atom electron is excited to an energy of −13.6/4 eV. How many different quantum states could the electron be in? That is, how many wave functions ψnℓm have this energy?

A.  1 B.  2 C.  3 D.  4 E.  more than 4

E = −13.6/4 eV means n2 = 4 so n = 2

For n = 2, ℓ = 0 or ℓ = 1.

For ℓ = 0, m = 0. For ℓ = 1, m = −1, 0, or 1 1 3


Degeneracy When multiple combinations of quantum numbers give rise to the same energy, this is called degeneracy. Ground state: n = 1, ℓ = 0, m = 0 no degeneracy

1st excited state: n = 2, ℓ = 0, m = 0 n = 2, ℓ = 1, m = −1 n = 2, ℓ = 1, m = 0 n = 2, ℓ = 1, m = 1

4 states (fourfold degenerate)

2nd excited state: n = 3, ℓ = 0, m = 0 n = 3, ℓ = 1, m = −1 n = 3, ℓ = 1, m = 0 n = 3, ℓ = 1, m = 1 9 states

(ninefold degenerate)

n = 3, ℓ = 2, m = −2

n = 3, ℓ = 2, m = 0 n = 3, ℓ = 2, m = 1 n = 3, ℓ = 2, m = 2

n = 3, ℓ = 2, m = −1

1s state

2s state

2p states

3s state

3p states

3d states


Hydrogen energy levels

n = 1

n = 2

n = 3

ℓ = 0 (s)

ℓ = 1 (p)

ℓ = 2 (d)

1s

2s 2p

3s 3p 3d


What do the wave functions look like?

ℓ (restricted to 0, 1, 2 … n-1) m (restricted to –ℓ to ℓ)

n = 1, 2, 3, …

2s

1s

3s 4s (ℓ=0) 4p (ℓ=1) 4d (ℓ=2)

Increasing n

Increasing ℓ 4f (ℓ=3, m=0)

m = −3

m = 3

Increases distance from nucleus, Increases # of radial nodes

Increases angular nodes Decreases radial nodes

Changes angular distribution


Radial part of hydrogen wave function Rnl(r) Radial part of the wave function for n=1, n=2, n=3.

Number of radial nodes (R(r) crosses x-axis or |R(r)|2 goes to 0) is equal to n−ℓ-1

x-axis is in units of the Bohr radius aB.

Quantum number m has no affect on the radial part of the wave function.


|Rnl(r)|2

Note that all ℓ=0 states peak at r=0

Since angular momentum is the electron cannot be at r=0 and have angular momentum.

Does this represent the probability of finding the electron near a given radius?

Not quite.

The radial part of the wave function squared


Clicker question 3 Set frequency to AD Assume that darts are thrown such that the probability of hitting any point is the same. The double ring is at r = 16.5 cm and the triple ring is at a r = 10.0 cm. Each ring has the same width in r. For a given dart, what is the probability of hitting a double compared to the probability of hitting a triple? That is, what is P(double)/P(triple)?

A.  1 B.  1.28 C.  1.65 D.  2.72 E.  Some other value


Clicker question 3 Set frequency to AD Assume that darts are thrown such that the probability of hitting any point is the same. The double ring is at r = 16.5 cm and the triple ring is at a r = 10.0 cm. Each ring has the same width in r. For a given dart, what is the probability of hitting a double compared to the probability of hitting a triple? That is, what is P(double)/P(triple)?

A.  1 B.  1.28 C.  1.65 D.  2.72 E.  Some other value

The width in r is the same (dr) so to get the area we multiply this width by the circumference (2πr).

So probability is proportional to r

Can also consider the differential area in polar coordinates

θ

θ


Probability versus radius: P(r) = |Rnl(r)|2r 2 In spherical coordinates, the volume element has an r2 term so probability increases with r2.

Most probable radius for the n = 1 state is at the Bohr radius aB

Most probable radius for all ℓ=n-1 states (those with only one peak) is at the radius predicted by Bohr (n2 aB).

Lecture 37 - University of Coloradojcumalat/phys2170_f13/lectures/Lec37.pdfwith the quantum numbers...

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