Download AP B - Unit 11 - 2013

AP Physics “B” - Unit 11 Notes – Yockers
Blackbody Radiation and Planck’s Hypothesis (pp. 874-877)
- an object at any temperature is known to emit electromagnetic radiation that is sometimes
referred to as thermal radiation
- thermally agitated charges emit radiation
→low temperatures → infrared radiation
→”warmer” → visible light
→high temperatures → ultraviolet radiation
- a blackbody is an ideal system that absorbs all radiation incident upon it
- the energy radiated from a blackbody varies with wavelength and temperature
→Wien’s displacement law
max T  0.2898 102 m  K
- max is the wavelength at which an energy curve peaks
- T is the absolute temperature of the object emitting the radiation
- the wavelength of electromagnetic waves radiated from an object is dependent solely on the
temperature of the object
- Planck – resonators (charged oscillators) only have certain discrete energies (ultraviolet
catastrophy)
En  nhf
→n = quantum number (positive integer)
→f = frequency of resonator’s oscillation
→h = 6.626 x 10-34 J∙s = Plank’s constant
- energy is quantized (the birth of quantum physics!)
Photoelectric Effect and the Particle Theory of Light (pp. 877-880)
- photoelectric effect – when light is incident on certain metallic surfaces, electrons
(photoelectrons) are emitted
→description of a photoelectric experiment/circuit
- maximum kinetic energy of the photoelectrons is related to stopping potential by
KEmax  eVs
- several features of the photoelectric effect cannot be explained with classical physics or with
the wave theory of light
→no electrons are emitted if the incident light frequency falls below some cutoff frequency, fc, that is
characteristic to the material being illuminated. This is inconsistent with the wave theory, which
predicts that the photoelectric effect should occur at any frequency, provided the light intensity is
sufficiently high
→the maximum kinetic energy of the photoelectrons is independent of light intensity. According to wave
theory, light of higher intensity should carry more energy into the metal per unit time and therefore
eject photoelectrons having higher kinetic energies
→the maximum kinetic energy of the photoelectrons increases with increasing light frequency. The wave
theory predicts no relationship between photoelectron energy and incident light frequency
-9
→electrons are emitted from the surface almost instantaneously (less than 10 s after the surface is
illuminated), even at low intensities. Classically, we expect the photoelectrons to require some
time to absorb some incident radiation before they acquire enough kinetic energy to escape from
the metal
- sketch or identify a graph of Vs versus frequency for a photoelectric experiment
→from such a graph determine threshold frequency and work function
→calculate an approximate value of h e
- Einstein (Nobel Prize in 1921) – a tiny packet of light energy or photon would be emitted
when a quantized oscillator made a jump from an energy state of En=nhf to the next
lower state En-1=(n-1)hf.
→photon – discrete unit of light emitted by a resonator
→conservation of energy would require the decrease in oscillator energy, hf, to be equal
to the photons energy, E, so
hc
E  hf 

- be able to describe in units of J or eV
- 1 eV = 1.60 x 10-19 J
→the maximum kinetic energy of “liberated” photoelectrons is
KEmax  hf  
- is called the work function of the metal which represents the minimum energy
with which an electron is bound in the metal (on the order of a few eV)
- with the photon theory of light, we can explain the previously mentioned features of the
photoelectric effect that cannot be understood using concepts of classical physics
→that the effect is not observed below a certain cutoff frequency follows from the fact that the photon
energy must be greater than or equal to . If the energy of the incoming photon does not satisfy
this condition, the electrons are never ejected from the surface, regardless of light intensity
→that KEmax is independent of the light intensity can be understood with the following argument. If the
light intensity is doubled, the number of photons is doubled, which doubles the number of
photoelectrons emitted. However, their maximum kinetic energy depends only on the light
frequency and the work function, not on the light intensity
→that KEmax increases with increasing frequency (KEmax = hf - )
→that the electrons are emitted almost instantaneously is consistent with the particle theory of light, in
which the incident energy arrives at the surface in small spatial packets and there is one-to-one
interaction between photons and photoelectrons. In this interaction, the photon’s energy is
imparted to an electron that then has enough energy to leave the metal. This is in contrast to the
wave theory, in which the incident energy is distributed uniformly over a large area of the metal
surface
Wave Properties of Particles (pp. 887-890)
- Louis de Broglie (Nobel Prize in 1929) postulated that because photons have wave ( and f)
and particle (mv and E) characteristics, perhaps all forms of matter have both properties
→the relationship between energy and momentum for a photon is p=E/c
→the momentum of a photon can be expressed as
E hc h
p 

c c 
→because the momentum of a particle of mass m and speed v is mv=p, the de Broglie
wavelength of a particle is (De Broglie’s hypothesis)
h
h
 
p mv
→de Broglie postulated that the frequencies of matter waves obey the Einstein
relationship E=hf, so
E
f 
h
- The Davisson-Germer Experiment (p. 902)
→provides evidence for the wave nature of electrons
→description (the discovery was made by accident!)
- involved the scattering of low-energy electrons (≈ 54eV) projected toward a
nickel target in a vacuum
- in one experiment the nickel target was badly oxidized because of an accidental
break in the vacuum system
- when the nickel target was heated in a flowing stream of hydrogen to remove
the oxide coating, electrons scattered by it exhibited intensity maxima and
minima at specific angles
- the nickel had formed large crystal regions on heating and the regularly spaced
planes of atoms in the crystalline regions served as a diffraction grating for
electrons
→Davisson and Germer performed more extensive diffraction measurements on
electrons scattered from single-crystal tagets
→results showed conclusively the wave nature of electrons
→confirmed the de Broglie relationship p  h 
X-rays (pp. 880-883)
- typical x-ray wavelengths are about 0.1 nm, which is on the order of the atomic spacing in a
solid
- x-rays are part of the electromagnetic spectrum
→frequencies higher than those of ultraviolet radiation
→can penetrate most materials with relative ease
- x-rays are produced when high-speed electrons are suddenly decelerated
→when a metal target is struck by electrons that have been accelerated through a
potential of several thousand volts
→electrons emit radiation when they undergo a deceleration inside the target
- as an electron passes close to a positively charged nucleus contained in the
target material, it is deflected from its path because of its electrical
attraction to the nucleus (deflection is due to acceleration)
- classical physics shows that any charged particle will radiate energy in the form
of electromagnetic radiation when it is accelerated
- quantum theory dictates that the radiation will be in the form of photons
- since the radiated photons carry energy, the electrons involved must have lost
kinetic energy during their encounters with the target nuclei
- in an extreme example when all of an electron’s energy is lost in a single “collision”
→the initial energy of the electron is transformed completely into the energy of the
photon
hc
eV  hf max 
min
→and the shortest wavelength that can be produced is
hc
min 
eV
- the x-ray spectrum (pp. 938-939) of a metal target consists of a broad continuous spectrum
(bremsstrahlung) plus a number of sharp lines, which are due to characteristic x-rays
→bremsstrahlung – x-radiation with its origin in the slowing down of electrons (braking
radiation)
→characteristic x-rays
- a bombarding electron collides with a target atom and has sufficient energy to
remove an inner-shell electron from the atom
- the vacancy created in the shell is filled when an electron in a higher level drops
down into the level containing the vacancy
- this transition is accompanied by the emission of a photon whose energy equals
the difference in energy between the two levels
→detailed x-ray spectra
Compton Scattering – Arthur Compton (p. 885-887)
- further justification for the photon theory of light
- description of the experiment
→x-ray beam of wavelength 0 directed toward a block of graphite
→scattered x-rays had a slightly longer wavelength,  , than the incident x-rays (lower
energy)
→remember, both energy and momentum must be conserved
- Compton shift formula
h
1  cos  
    0 
me c
h
→
is known as the Compton wavelength and has a value of 0.00243 nm
me c
→the Compton shift depends on the scattering angle,  , and not on wavelength
Atomic Spectra (pp. 904-910)
- when an evacuated glass tube is filled with a gas at very low pressure with a potential
difference applied between metal electrodes in the tube great enough to produce an
electric current in the gas, the tube emits light whose color is characteristic of the gas in
the tube
- when the emitted light is analyzed with a spectrometer
→a series of discrete lines are observed (emission spectrum)
→each line corresponds to a different wavelength (color) of light
- the wavelengths contained in a given line spectrum are characteristic of the element emitting
the light
- because no two elements emit the same line spectrum, this phenomenon provides a
technique for identifying elements in a substance
- an element can also absorb light at specific wavelengths with corresponding spectral lines
giving an absorption spectrum
→an absorption spectrum is obtained by passing a continuous radiation spectrum
through a vapor of the element being analyzed
→the absorption spectrum consists of a series of dark lines superimposed on the
otherwise continuous spectrum
→each line in the absorption spectrum of a given element coincides with a line in the
emission spectrum of the element
- radiation is emitted by an atom when an electron “jumps” from a more energetic initial state to
a lower state
→frequency of the emitted radiation is
Ei  E f  hf
→to emit radiation Ei  E f
- energy incident upon an atom causes an electron to “jump” from a lower energy state to a
more energetic state
→energy is absorbed by the atom
→”excited” atoms quickly fall to lower energy state – this emits radiation as described
above
- to determine the wavelength of emitted radiation
 1
1
1 
 RH  2  2 
n


 f ni 
→the Rydberg constant – RH = 1.0973732 x 107 m-1
→n has integral values corresponding to energy levels
- to determine the energies of the quantum levels (for hydrogen)
13.6
E n   2 eV
n
- the minimum energy required to ionize the atom (to completely remove the electron in
hydrogen, for example) is called the ionization energy
- in the diagram above, the uppermost level shown, corresponding to n   , represents the
state for which an electron is completely removed
→the ionization energy for hydrogen is 13.6 eV
→the wavelength required to ionize an atom can be determined by using the equations
above for wavelength or frequency (and the conversion from J to eV)
- the various series in the hydrogen spectrum
→Paschen series – infrared
→Balmer series – visible
→Lyman series – ultraviolet
- example
The Balmer series for the hydrogen atom corresponds to electronic transitions that
terminate in the state with quantum number n=2, as shown above. (a) Find the longestwavelength photon emitted in the Balmer series and determine its frequency and energy.
(b) Find the shortest-wavelength photon emitted in the same series.
 1
1
1 
13.6
Ei  E f  hf
 RH  2  2 
E n   2 eV
n


n
 f ni 
Properties of Nuclei (p. 940-976)
- nucleons
→protons
→neutrons
- symbols
→mass number (A) = the number of nucleons in the nucleus
→neutron number (N) = number of neutrons in the nucleus
→atomic number (Z) = number of protons in the nucleus
→ ZA X is the symbol used to represent nuclei
- X represents the chemical symbol for the element
- isotopes of an element have the same Z value but different N and A values
- charge and mass (p. 940)
→charge of a proton = +e
→charge of an electron = -e
→neutrons are electrically neutral
- nuclear stability (p. 942) – the nuclear force
→the very large repulsive electrostatic forces between protons should cause the
nucleus to fly apart
→nuclei are stable because of the presence of a short-range force – nuclear force
- acts over distances of about 2 fm (f = femto → 10-15)
- acts between all nuclear particles
→neutron-neutron
→proton-neutron
→proton-proton
- protons attract each other via the nuclear force and at the same time repel each
other through the Coulomb force
- nuclear force dominates the Coulomb repulsive force within the nucleus (at
short ranges)
- the nuclear forces associated with neutron-neutron, proton-neutron, and proton
proton interactions are approximately the same apart from the additional
repulsive Coulomb force for the proton-proton interaction
- the decay processes (pp. 948-952) – radioactive nuclei spontaneously decay via alpha, beta,
and gamma decay
→alpha decay
- if a nucleus emits an alpha particle 42 He it loses two protons and two neutrons
→N decreases by 2
→Z decreases by 2
→A decreases by 4
A
A 4
4
Z X Z  2Y 2 He
→X is called the parent nucleus
→Y is the daughter nucleus
- examples
238
234
4
92 U 90Th 2 He

226
88

4
Ra  222
86 Rn  2 He
http://phet.colorado.edu/en/simulation/alpha-decay
→beta decay – the daughter nucleus has the same number of nucleons as the parent
nucleus, but the atomic number is changed by 1
A
A

Z X Z 1Y  e
(e+ is a positron – antiparticle of an electron)
A
A

Z X Z 1Y  e
- nucleon number and total charge are conserved
- example: 146 C147Y  e 
- where did the electron come from (this is the nucleus)?
→the emitted electron is created in the nucleus by a process in which a
neutron is transformed into a proton  01 n11 p  e  
- beta decay posed a problem because it appeared to violate the conservation of
energy and momentum
→a third particle must be present to carry away the “missing” energy and
to conserve momentum
→the neutrino   was finally discovered in 1950
- no electric charge
- mass smaller than an electron (possibly zero!)
- has spin
- weak interactions with matter – hard to detect
- correct equations with the addition of neutrinos
A
A

Z X Z 1Y  e  
X ZA1Y  e   
- in beta decay, an electron and an antineutrino are emitted or a positron and a
neutrino
http://phet.colorado.edu/en/simulation/beta-decay
→gamma decay
- a nucleus may reach an excited state as the result of a violent collision with
another particle
- more common for a nucleus to be in an excited state as a result of alpha or beta
decay
- gamma decay is a “second” decay for the excited nucleus to reach a lower
energy state by emitting one or more photons (similar to the emission of
light by an atom)
→photons emitted are gamma rays
→gamma rays have very high energy relative to the energy of visible light
- example
12
12 *

5 B 6 C  e
A
Z
C* 126 C  
→the * indicates that the carbon nucleus is left in an excited state
following the decay
- gamma emission does not result in any change in either Z or A
- nuclear fission (p. 973-976) – occurs when a heavy nucleus splits, or fissions, into two
smaller nuclei
→the total mass of the products is less than the original mass of the heavy nucleus
→the fission of 235 U by slow (low energy) neutrons can be represented by the reaction
1
235
236 *
0 n  92 U 92 U  X  Y  neutrons X and Y are known as fission fragments
12
6
- the 235 U nucleus captures a thermal (slow-moving) neutron
- the capture results in the formation of 236 U * , and the excess energy of this
nucleus causes it to undergo violent oscillations
236 *
- the U nucleus becomes highly distorted, and the force of repulsion between
protons in the two halves of the dumbbell shape tends to increase the
distortion
- the nucleus splits into two fragments, emitting several neutrons in the process
→typical reaction of this type is
1
235
141
92
1
0 n  92 U 56 Ba  36 Kr 30 n
- the fission fragments, barium and krypton (in above example), and the released
neutrons have a great deal of kinetic energy
- the amount of energy released is (8.2 – 7.2) MeV per fission event, so if we
assume a total of 240 nucleons
MeV
MeV 

Q  240 nucleons  8.2
 7.2
  240 MeV
nucleon
nucleon 

→Q is the disintegration energy for a fission process
http://phet.colorado.edu/en/simulation/nuclear-fission
- using E  mc 2 to calculate energy released in nuclear processes
→example
- 146 C atom147 N atom
- 146 C has a mass of 14.003242 u and 147 N has a mass of 14.003074
- m  0.000168 u
- this corresponds to an energy release of
E  mc 2  0.000168 u 931 MeV u   0.156 MeV


E  mc 2  0.000168 u  1.66  10 27 kg u 3.00  108 m s
931 M eV
c2 
u

931 M eV 1.60  10 -19 J eV
c
u 1.66  10 -27 kg u

c  3.00  10 8 m s



2
 2.51 10 14 J