WileyPLUS Assignment 3

WileyPLUS Assignment 3

Chapters 22, 24 - 27

Due Friday, March 27 at noon

Q24.21: Intensity = Power/AreaQ25.10: Concave mirror - not covered in course, do not attemptQ26.72: Farsighted person - incorrectly marked sometimes?

Assignment 4 to follow, due last day of term

1Friday, March 27, 2009

Week of March 30 - April 3

Tutorial and Test 4: ch. 27, 28

PHYS 1030 Final Exam

Wednesday, April 15

9:00 - 12:00

Frank Kennedy Gold Gym

30 multiple choice questionsFormula sheet provided


Then, !y !py ! h

Heisenberg Uncertainty Principle

sin!= "/W

Diffraction minima:W sin!= m"

So, !py !"

W× px !

"

W× h

"

!py !h

W

m = 1

m = 1

or, W!py ! h

Heisenberg: sending the particle through the aperture is a way of measuring the particle’s position in the y-direction to a precision: Δy = W.

For small angles, !! "

W! #py

px

m = 1

! = h/px


Heisenberg Uncertainty Principle – Alternative View

The position of a particle is measured by shining light on it. Because the photons carry momentum, the act of observing the particle perturbs its motion so that its momentum becomes uncertain.

The ultimate resolution in measuring the position of the particle is limited by the wavelength of the light, so "y ! !.

The photon carries momentum and disturbs the motion of the particle in bouncing off it. So the momentum of the particle after observing it is uncertain to "py ! p = h/!.

Then, "y !py ! λ " h/! = h

vy

Photon,

p = h/!

Reflected

photon

Particle


Because of the wavelike behaviour of matter it is not possible to measure both the position and momentum of an object to arbitrarily high precision.

A more complete analysis shows that the best possible measurements of position and momentum follow the Heisenberg uncertainty relation:

!y !py ≥h

4"

A similar relation holds for the uncertainty of the energy of a particle and the time during which the energy is measured or during which the particle occupies an energy state:

!E !t ≥ h

4"

The Heisenberg uncertainty relations apply to any measurements.

Heisenberg Uncertainty Principle

Also, ∆x ∆px ≥h

4π, ∆z ∆pz ≥

h4π


A source of uncertainty...

Edition 5 of Cutnell defines the uncertainty relation as:

!y !py ≥h

2"

We use the version in editions 6 and 7:

!E !t ≥ h

4"!y !py ≥

h

4"

The formula sheet uses this version


Prob. 29.32: An electron is trapped within a sphere whose diameter is 6"10-15 m (~size of the nucleus of an oxygen atom).

What is the minimum uncertainty in the electron’s momentum?


Prob. 29.-/36: Suppose the minimum uncertainty in the position of a particle is equal to its de Broglie wavelength.

If the particle has an average speed of 4.5"105 m/s, what is the minimum uncertainty in its speed?


Summary, Wave-Particle Duality

• Light has properties of particles: – energy is carried in packets of radiation (photons) of energy hf = hc/! (photoelectric effect, blackbody radiation) – Photons carry momentum p = E/c = h/! (verified by Compton effect, not covered)

• Particles have properties of waves: – diffraction according to a de Broglie wavelength ! = h/p

• Energy does not take on just any value, but is quantized

" (Planck’s theory of blackbody radiation). Further evidence in quantization of energy levels of atoms (next chapter).

• There is a fundamental limit to how well both position and momentum can be known (Heisenberg Uncertainty Principle).


Chapter 30: The Nature of the Atom

• Rutherford scattering and the nuclear atom - –#the nucleus is really small!

• Atomic line spectra – wavelengths characteristic of each element

• Bohr model of the hydrogen atom –# - quantization of angular momentum ! quantization of energy

• x-rays

• The laser

• Omit 30.5, 6 – quantum mechanical picture of H-atom, Pauli Exclusion Principle, periodic table


Early model of the atom

• About 10-10 m in size.

• Positively charged (“pudding”), with negatively charged electrons (“plums”) embedded in the pudding – “plum pudding model”

Stability of the plum pudding was questionable.

Where would the characteristic spectral lines come from?

#


Geiger+Marsden: Fall of the plum puddingAlpha (#) particles – nuclei of 4He atom, emitted by some radioactive nuclei – were scattered from a thin gold foil and observed on a screen as flashes of light.

Far more were scattered at large angle than would be possible with the weak electric field inside a “plum pudding” atom.

Rutherford: realized that the positive charges of the atom had to be contained in a very small volume – the nucleus –#electric field very strong close to it.

$ planetary model of the atom with electrons in orbit around nucleus


Very schematic picture of an atom

Size of the nucleus ! 10-15 – 10-14 m

Size of the atom ! 10-10 m

Equal amounts of + and – charges in the neutral atom.

Electrons are in orbit around the nucleus in a planetary model of the atom.

(charge +Ze)


Problem with a Planetary Model of the Atom

• The electrons suffer centripetal acceleration in their orbital motion.

• Accelerated charges should radiate electromagnetic energy.

$ The electrons should lose energy and spiral into the nucleus in very little time.

$ A planetary atom should not be stable!

$ Classical theory does not explain the structure of the atom.

$ Small systems, such as atoms, must behave differently from

large.


Fraunhofer absorption lines from the sun

http://www.harmsy.freeuk.com/fraunhofer.html

1 Å (Angstrom) = 0.1 nm

• Fraunhofer lines – due to absorption of sunlight by elements in the atmosphere of the sun •chemical elements emit and absorb light at the same wavelengths• Models of the atom need to explain this

The sun: blackbody radiation for T ! 6000 K


Line spectrum of the hydrogen atom

InfraredUltraviolet Visible

m Series

1 Lyman ultraviolet

2 Balmer visible

3 Paschen infrared

4 Brackett infrared

Balmer found by trial and error a simple formula to calculate the wavelength of all lines of the hydrogen atom:

R = Rydberg constant = 1.097 " 107 m-1

1

!= R

(1

m2− 1

n2

)m= 1,2,3 . . .

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

n = 4n = 5

n = 6

n = $

series limit

n = 3

n = 4n = 2

m = 1 m = 2 m = 3+ other series


1

!= 1.097×107

(1

12− 1"

)= 1.097×107

! = 91.2 nm (series limit)

Spectrum of H-atom

Example: Lyman series, m = 1:

For n = $: “series limit”

For n = 2:

1

!= 1.097×107

(1

12− 1

22

)= 0.823×107

! = 121.5 nm

R = Rydberg constant = 1.097 " 107 m-1

1

!= R

(1

m2− 1

n2

)m= 1,2,3 . . .

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

(series limit, shortest wavelength in Lyman series)

(longest wavelength in Lyman series)

m = 1

n = 2n = $

m–1

m–1


Bohr model of the hydrogen atom

Now superseded by more modern quantum mechanical ideas, but gives the correct answers.

Assume:• A planetary model with electron in orbit around the nucleus.• There are certain electron orbits that are stable (“stationary states”). This is contrary to classical theory and is not explained in this model.• Light is emitted or absorbed when an electron changes state.• Energy is conserved, so the energy of the photon is the difference in energy between the initial and final states:

E = hf = Ei – Ef " – what are the allowed energies?

E = hf

“Stationary” states

= Ei – Ef


The electron is kept in orbit by the attractive Coulomb force between it and the nucleus.

F =kZe

2

r2=mv

2

r

mv2 =

kZe2

r

The potential energy is:

PE=−kZe2

r

And so the total mechanical energy, E = KE + PE, is:

But, what are the “good” values of r corresponding to the stationary states?

= 2 KE (A)

E =kZe

2

2r− kZe

2

r=−kZe

2

2r(B)

Energies of hydrogen atom in the Bohr model

A hydrogen-like atom or ion with just one electron in orbit

m


Combine with (A): mv2r = kZe

2

So, v2 =

kZe2

mr=

[nh

2!mr

]2

r =1

kZe2m

[nh

2!

]2= (5.29×10−11 m)

n2

Z

En =−[2!2mk2e4

h2

]Z2

n2, n= 1,2,3 . . .

Ln = mvr =nh

2!n = 1, 2, 3...

Energies of hydrogen atom in the Bohr model

And then, substituting into (B), the total mechanical energy is:[E =−kZe

2

2r

]

En =−13.6 Z2

n2eV

Assertion (Bohr): the angular momentum of the electron around the nucleus can have only certain, quantized, values:

m

n = “quantum number”


Emission of a photon

Emission and absorption occur at the same wavelengths (as seen in the Fraunhofer absorption lines of sunlight).

E = hfEi = Ef +h f

Absorption of a photon

E = hf

Ef +h f = Ei


Spectrum of hydrogen-like atoms and ions

Energy levels: En =−13.6 Z2

n2eV

E = hf

ni

nf

(Just one electron in orbit around the

nucleus)

In agreement with Balmer’s formula[R=

(13.6×1.602×10−19 J)hc

= 1.097×107 m−1]

= Ei – E

f

Energy of photons: E = Ei − Ef =hc

λ= 13.6Z2

[1n2

f

− 1n2

i

]eV

Z = 1 for hydrogen


Energy levels of the hydrogen atom – Bohr

En

erg

y %

En

erg

y %

Origin of the lines of the hydrogen atom spectrum

En =−13.6n2

eV

Ground state

Free electron

(Z = 1)


Prob. 30.-/7: It is possible to use electromagnetic radiation to ionize atoms. To do so, the atoms must absorb the radiation, the photons of which must have enough energy to remove an electron from an atom.

What is the longest radiation wavelength that can be used to ionize the ground state of the hydrogen atom? En

ergy

→

Free electron

The least energy to remove a ground state electron is E$ – E1.

Ground state,n = 1


Ener

gy →

Ground state

Free electron

First excited state, n = 2

Prob. 30.7/8: The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy.

What is the quantum number, n, of the state into which the electron moves?


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WileyPLUS Assignment 3