modern physics. what is meant by “quantized”? what is meant by “quantized”? – quantity...

Modern Physics

• What is meant by “quantized”?–Quantity• Specific and discrete quantity • Packets of definite size.

• Quantized energy can be thought of as existing in very small packets of specific size.

• Atoms absorb and emit quanta of energy.

• Let us consider light…

Electromagnetic Spectrum

Visible spectrum

Three models are used to describe light• What model is used in geometric optics, like with

lenses and mirrors?–Ray

• What model is used in studying diffraction and interference?–Wave

• What model is used to study interaction of light and atoms?–Particle ( photon)

But light is said to have a “dual nature”• What is that supposed to mean?– Wave particle duality • Waves have both a wave and a particle component• We describe the path of light as a ray

– Equation• E = hf for a single photon• E = nhf for multiple photons–h = plank’s constant»6.63x10-34 J s ∙ (SI version)»4.14x10-15 eV s ∙ (convenient)

–f = frequency–n = number of photons

Conceptual checkpoint• Which has more energy in its photons, a very bright,

powerful red laser or a small key-ring red laser?–Neither! They both have the same energy per

photon. The big one has more power.

• Which has more energy in its photons, a red laser or a green laser?– The green one has shorter wavelength and higher

frequency. It has more energy per photon.

The “electron-volt” (eV)• The electron-volt is the most useful unit on the

atomic level.• If a moving electron is stopped by 1 V of electric

potential, we say it has 1 electron-volt (or 1 eV) of kinetic energy.

• 1 eV = 1.602×10-19 J

• What is the frequency and wavelength of a photon whose energy is 4.0 x 10-19J?

E = hff = E / h = 4.0x10-19 J / 6.625x10-34 J s∙

E = 4.0x10-19 J h= 6.625x10-34 J s∙

= 6.04x1014 Hz

λ = c/f

λ = c/f = 3x108 m/s / 6.04x1014 1 / s = 4.97x10-7 m

= 497 nm

• How many photons are emitted per second by a He-Ne laser that emits 3.0 mW of power at a wavelength of 632.8 nm?

E =n(hf)

f = c / λ

P = E /t

P = 3.0 mW = 0.003 W λ = 632.8 nm = 632.8x10-9 m

E = P t∙ E = 0.003 W 1 s∙ = 0.003 J

n =E / (hf)

= 3x108 m/s / 632.8x10-9 m = 4.74x1014 Hz

=0.0003 J / (6.625x10-34 J s 4.74x10∙ ∙ 14 Hz) = 9.55x1014

• What are atoms composed of?– Atoms consist of nuclei (protons and neutrons and

electrons.• What happens when an atom encounters a photon?– The atom usually ignores the photon, but sometimes the

atom absorbs the photon.• If the photon is absorbed by the atom, what

happens next?– The photon disappears and winds up giving all its energy

to the atom’s electrons.

• This is a graph of energy levels for a hypothetical atom

0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV

Ionization levelThird excited state

Second excited state

First excited state

Ground state (lowest energy level

Atom loses an electron

Highest energy level

Higher energy levels. (Atoms has to absorb energy to get from the ground state.

Normal “unexcited” state

• What do we mean when we say the atoms energy levels are “quantized”?

0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state


•Only certain energies are allowed.•These are represented by the horizontal lines.•The atom cannot exist at energies not shown in this graph!

Absorption of photon by Atom • When a photon of light is absorbed by an atom, it

causes an increase in the energy of the atom.• The photon disappears, and the energy of the

atom increases by exactly the amount of energy contained in the photon.

• The photon can be absorbed ONLY if it can produce an “allowed” energy increase in the atom.

• Absorption of photon by atom

Exited state

-10.0eV Ground state

ΔE = hf E = hf

0.0eV

Absorption Spectrum• When an atom absorbs photons, it removes the

photons from the white light striking the atom, resulting in dark bands in the spectrum.

• Therefore, a spectrum with dark bands in it is called an absorption spectrum.

• Absorption Spectrum

0.0eV


Absorption spectra always involve atoms going up in energy level.

Ionized

Emission of photon by atom• When a photon of light is emitted by an atom, it

causes a decrease in the energy of the atom.• A photon of light is created, and the energy of the

atom decreases by exactly the amount of energy contained in the photon that is emitted.

• The photon can be emitted ONLY if it can produce an “allowed” energy decrease in an excited atom.

• Emission of photon by atom

0.0eV


ΔE = hf E = hf

Exited state

Emission Spectrum• When an atom emits photons, it glows! The

photons cause bright lines of light in a spectrum.• Therefore, a spectrum with bright bands in it is

called an emission spectrum.

• Emission of photon by atom

0.0eV


Emission spectra always involve atoms going down in energy level.

Ionized

Photoelectric effect

What is the frequency and wavelength of the light that will cause the atom shown to transition from the ground state to the first excited state? Draw the transition.

0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state


ΔE = hf f = Δ E / h

f = (11.5 – 3.0) / 4.14x10-15

h = 4.14x10-15 eV s∙

f = 1.45x1015 Hz

λ= c / f

λ= 3x108 / 1.45x1015 λ= 2.07x10-7 m

λ= 207 nm

What is the longest wavelength of light that when absorbed will cause the atom shown to ionize from the ground state? Draw the transition.

0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state



f = (11.5 ) / 4.14x10-15

h = 4.14x10-15 eV s∙

f = 2.78x1015 Hz

λ= c / f

λ= 3x108 / 2.78x1015 λ= 1.08x10-7 m

λ= 1-8 nm

The atom shown is in the second excited state . What frequencies of light are seen in its emission spectrum?Draw the transition.

0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state



f = (11.5 – 3.0 ) / 4.14x10-15

h = 4.14x10-15 eV s∙

f = 2.053x1015 Hz1 2

3

1


0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state



f = (5.5-3 ) / 4.14x10-15

h = 4.14x10-15 eV s∙

f = 6.09x1014 Hz1 2

3

2


0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state



f = (11.5-5.5 ) / 4.14x10-15

h = 4.14x10-15 eV s∙

f = 1.45x1015 Hz1 2

3

3


0.0eV-1.0eV

-3.0eV

-5.5 eV

-11.5 eV



First excited state


1 2

3

2f = 6.09x1014 Hz

1f = 2.053x1015 Hz

3f =1.45x1015 Hz

Atoms absorbing photons increase in energy

0.0eV

-4.0eV

-12 eV

Ionization level


•We’ve seen that if you shine light on atoms, they can absorb photons and increase in energy.•The transition shown is the absorption of an 8.0 eV photon by this atom.•You can use Planck’s equation to calculate the frequency and wavelength of this photon.

Photon - 4.0 eV with largest allowed energy

Question• Now, suppose a photon with TOO MUCH ENERGY

encounters an atom?• If the atom is “photo-active”, a very interesting and

useful phenomenon can occur…• This phenomenon is called the

Photoelectric Effect.

Photoelectric Effect

0.0eV

-4.0eV

-12 eV

Ionization level


•Some “photoactive” metals can absorb photons that not only ionize the metal, but give the electron enough kinetic energy to escape from the atom and travel away from it. •The electrons that escape are often called “photoelectrons”.•The binding energy or “work function” is the energy necessary to promote the electron to the ionization level.•The kinetic energy of the electron is the extra energy provided by the photon.

W0 = Work function

Kinetic energy

e-

Phot

on e

nerg

y

Eph

E = W0 + KE

Photoelectric Effect

0.0eV

-4.0eV

-12 eV

Ionization level


• Photon Energy = Work Function + Kinetic Energy • hf = Φ + Kmax

• Kmax = hf – Φ• K:max Kinetic energy of

“photoelectrons” • hf: energy of the

photon• Φ: binding energy or

“work function” of the metal.

W0 = Work function

Kinetic energy

e-

Phot

on e

nerg

y

Eph

E = W0 + KE

• Suppose the maximum wavelength a photon can have and still eject an electron from a metal is 340 nm. What is the work function of the metal surface?

The longest wavelength is the lowest energy, and will provide no “extra” kinetic energy for the electron.

Kmax = hf – Φ 0 J = hf – Φ

Φ = hf Φ = hv / λ f = v / λ

Φ = (4.14x10-15eV 3x10∙ 8 m/s) / 340x10-9 m Φ = 3.65 eV

Question• Suppose you collect Kmax and frequency data for a

metal at several different frequencies. You then graph Kmax for photoelectrons on y-axis and frequency on x-axis. What information can you get from the slope and intercept of your data?Slope: Planck’s ConstantIntercept: Φ - binding energy or “work function”

The Photoelectric Effect experiment• The Photoelectric Effect experiment is one of the

most famous experiments in modern physics.• The experiment is based on measuring the

frequencies of light shining on a metal (which is controlled by the scientist), and measuring the resulting energy of the photoelectrons produced by seeing how much voltage is needed to stop them.

• Albert Einstein won the Nobel Prize by explaining the results.

Photoelectric Effect experiment

V

A

Metal (+)

CollectorLight

e- e- e- e- e- e-e-e-

e-

e-

e-

e-

e-e-e-e-e-e-e-

e-

e-

e-

Strange results in the Photoelectric Effect experiment • Voltage necessary to stop electrons is independent of

intensity (brightness) of light. It depends only on the light’s frequency (or color).

• Photoelectrons are not released below a certain frequency, regardless of intensity of light.

• The release of photoelectrons is instantaneous, even in very feeble light, provided the frequency is above the cutoff.

Voltage versus current for different intensities of light.

Number of electrons (current) increases with brightness, but energy of electrons doesn’t!

“Stopping Voltage”

Vs, the voltage needed to stop the electrons, doesn’t change with light intensity. That means the kinetic energy of the electrons is ndependent of how bright the light is.

Voltage versus current for different frequencies of light.

Energy ofelectronsincreases asthe energy ofthe lightincreases.

“Stopping Voltage”

Vs changes with light frequency. That means the kinetic energy of the photoelectrons is dependent on light color.

Experimental determination of the Kinetic Energy of a photoelectron

• The kinetic energy of photoelectrons can be determined from the voltage (stopping potential) necessary to stop the electron.

• If it takes 6.5 Volts to stop the electron, it has 6.5 eV of kinetic energy.

Graph of Photoelectric Equation

Kmax

f

Slope = h

Cut-off Frequency

Φ (binding energy)

KMAX = hf - Φy= mx + b

• Question: Does a photon have mass?

–A photon has a fixed amount of energy (E = hf).

–We can calculate how much mass would have to be destroyed to create a photon (E=mc2).

• Calculate the mass that must be destroyed to create a photon of 340 nm light.

E = hf

hf = mc2

λ = 340x10-9 m h = 6.63x10-34 J s∙

E = mc2 f = c / λ

hc / λ = mc2

h / λ = mcm = h / (λc)

m = 6.63x10-34 / (340x10-9 3x10∙ 8 )

h = 6.63x10-34 (kg m∙ 2 / s2) s∙

=

Momentum of a Photon• Does a photon have momentum?

Yes• A photon’s momentum is calculated by

–p = E / c = hf / c = h / λ

We have experimental proof ofthe momentum of photons

• Compton scattering–Proof that photons have momentum.–High-energy photons collided with electrons

exhibit conservation of momentum.–Work Compton problems just like other

conservation of momentum problems – except the momentum of a photon uses a different equation.

• What is the frequency of a photon that has the same momentum as an electron with speed 1200 m/s?

p = hf / c

melectron = 9.11x10-31 Kg

p = mv

p = 9.11x10-31 Kg 12 m / s∙ =

f = P c / hf = ( kg m/s) (3x10∙ ∙ 8 m/s) / (6.63x10-34 (kg m∙ 2 / s2) s)∙

h = 6.63x10-34 J s∙h = 6.63x10-34 (kg m∙ 2 / s2) s∙

f =

Wave-Particle Duality• Waves act like particles sometimes and particles

act like waves sometimes.• This is most easily observed for very energetic

photons (gamma or x-Ray) or very tiny particles (elections or nucleons)

Particles and Photons both have Energy

• A moving particle has kinetic energy– E = K = ½ mv2

• A particle has most of its energy locked up in its mass.– E = mc2

• A photon’s energy is calculated using its frequency– E = hf

Particles and Photons both have Momentum• For a particle that is moving– p = mv– kg m/s∙

• For a photon– p = h/λ– (kg m∙ 2 / s2) s / m = kg m / s∙ ∙– Check out the units! They are those of momentum.

Particles and Photons both have a Wavelength• For a photon– l = c/f

• For a particle, which has an actual mass, this equation still works– λ= h/p where p = mv–This is referred to as the deBroglie wavelength

We have experimental proof that particles have a wavelength

• Davisson-Germer Experiment–Verified that electrons have wave properties by

proving that they diffract.– Electrons were “shone” on a nickel surface and

acted like light by diffraction and interference.–We’ll study diffraction in the next unit, and

return to this experiment then…

• What is the momentum of photons that have a wavelength of 620 nm?

p = h / λ

λ = 620x10-9 m h = 6.63x10-34 J s∙

= 6.63x10-34 J s / 620x10∙ -9 m =

• What is the wavelength of a 2,200 kg elephant running at 1.2 m/s?

= 2200 kg 1.2 m/s∙

p = h / λ

m = 2200 kg v = 1.2 m / s

p = mv

= 6.63x10-34 J s / 12 m / s∙

=

λ = h / p =

Naming a Nucleus

C12

6

Mass number

Atomic number

• What are isotopes?• What characteristics do isotopes of the same

element share?

What characteristics do isotopes of the same element share?Atomic number

What characteristics do isotopes of the same element not share?Atomic massRadioactivity

Isotopes• Isotopes have the same atomic number and

different atomic mass.• Isotopes have similar or identical chemistry.• Isotopes have different nuclear behavior.• Examples:

Naming a Nucleus

126 C 13

6 C 146 C

Elementary Particles

1

1

1

0

0

-1

p

ne

0

+1 e

mass

charge mass

charge mass

charge

Negative chare Positive chare

Proton

Neutron

Electron

Nuclear reactions• Nuclear Decay: a spontaneous process in which an

unstable nucleus ejects a particle and changes to another nucleus.–Alpha decay–Beta decay• Beta Minus• Positron

• Fission: a nucleus splits into two fragments of roughly equal size.

• Fusion: Two nuclei combine to form another nucleus.

Decay Reactions• Alpha decay–A nucleus ejects an alpha particle, which is just a

helium nucleus.• Beta decay–A nucleus ejects a negative electron.

• Positron decay–A nucleus ejects a positive electron.

• Simulations– http://library.thinkquest.org/17940/texts/radioactivity

/radioactivity.html

http://library.thinkquest.org/17940/texts/radioactivity/radioactivity.html

http://library.thinkquest.org/17940/texts/radioactivity/radioactivity.html

Alpha (α) Decay

Alpha particle (helium nucleus) is released. Alpha decay only occurs with very heavy elements.

239

94Pu +

235

92U

4

2He

Beta (β-)DecayA beta particle (negative electron) is released. Beta decay occurs when a nucleus has too many neutrons for the protons present. A neutron converts to a proton. An antineutrino is also released.

14

6 C +

14

7N

0

-1e + γ

Beta (β+)DecayPositron (positive electron) is released. Positron decay occurs when a nucleus has too many protons for the neutrons present. A proton converts to a neutron. A neutrino is also released.

2

2 He +

2

1H

0

1e + γ

Neutrino and Anti-Neutrino• Proposed to make beta and positron decay obey

conservation of energy.• These particles possess energy and spin, but do

not possess mass or charge.• They do not react easily with matter, and are

extremely hard to detect.

Gamma Radiation, γ

•Gamma radiation is electromagnetic in nature.•Gamma photons are released by atoms which have just undergone a nuclear reaction when the excited new nucleus drops to its ground state.•The high energy in a gamma photon is calculated by E = hf.

• Complete the reaction, identify the type of decay.

234

90 Th +234

91 Pa 0-1

e + γ

• Complete the reaction for the alpha decay of Thorium-232

232

90 Th +

228

88Ra

4

2He

Nuclear Fission and Fusion

Fission

•Fission occurs when an unstable heavy nucleus splits apart into two lighter nuclei, forming two new elements. •Fission can be induced by free neutrons. •Mass is destroyed and energy produced according to E = mc2.• http://library.thinkquest.org/17940/texts/fission/fission.html• http://www.atomicarchive.com/Movies/index.shtml

http://library.thinkquest.org/17940/texts/fission/fission.html

http://library.thinkquest.org/17940/texts/fission/fission.html

http://www.atomicarchive.com/Movies/index.shtml

Neutron-induced fission• Neutron-induced fission produces a “chain

reaction.” What does that mean?• Nuclear power plants operate by harnessing the

energy released in fission in by controlling the chain reaction.

• Nuclear weapons depend upon the initiation of an uncontrolled fission reaction.

Critical Mass• The neutrons released from an atom that has

undergone fission cannot immediately be absorbed by other nearby fissionable nuclei until they slow down to “thermal” levels.

• How can this concept be used to explain why a chain reaction in nuclear fission will not occur unless a “critical mass” of the fissionable element is present at the same location?

Nuclear Reactors

•Nuclear reactors produce electrical energy through fission.•Advantages are that a large amount of energy is produced without burning fossil fuels or creating greenhouse gases.•A disadvantage is the production of highly radioactive waste.•Another simulation appears athttp://www.howstuffworks.com/nuclear-power.htm

http://www.howstuffworks.com/nuclear-power.htm

Nuclear Weapons

•Nuclear weapons have been used only twice, although they have been tested thousands of times.•Weapons based on nuclear fission involve slamming together enough material to produce an uncontrolled fission chain reaction.

Little Boy was dropped on Hiroshima and contained U-235 produced in Oak Ridge, TN.

Fission• Fission occurs only with very heavy elements,

since fissionable nuclei are too large to be stable.• A charge/mass calculation is performed to balance

the nuclear equation.• Mass is destroyed and energy produced according

to E = mc2.

Fusion• Fusion occurs when two light nuclei come

together to form a new nucleus of a new element.• Fusion is the most energetic of all nuclear

reactions.• Energy is produced by fusion in the sun.• Fusion of light elements can result in non-

radioactive waste.

Fusion

•Fusion is the reaction that powers the sun, but it has not been reliably sustained on earth in a controlled reaction.•Advantages to developing controlled fusion would be the tremendous energy output and the lack of radioactive waste products.•Disadvantages are – we don’t know if we’ll be technically able to do it on earth!

Mass defect• This strange term is used to indicate how much

mass is destroyed when a nucleus is created from its component parts.

• The mass defect is generally much, much less than the mass of a proton or neutron, but is significant nonetheless.

• The loss of mass results in creation of energy, according to E = mc2.

modern physics. what is meant by “quantized”? what is meant by “quantized”? – quantity...

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