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12. Metal, Semiconductors, LED (42.5-7), Nuclear Physics: Radioactivity (43.1-2)
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12. Metal, Semiconductors, LED (42.5-7), Nuclear Physics: Radioactivity (43.1-2)
Due: 11:59pm on Sunday, April 24, 2016
To understand how points are awarded, read the Grading Policy for this assignment.
± The Fermi Energy in Sodium
Description: ± Includes Math Remediation. Using the Fermi energy, calculate the average energy and speed of an electron at absolute zero in sodium metal. Find the
Fermi temperature for sodium.
The Fermi energy of sodium is 3.23
.
Part A
Find the average energy
of the electrons in sodium at absolute zero.
Express your answer in electron volts using three significant figures.
Hint 1. Average free-electron energy
For metals at absolute zero, the Fermi-Dirac distribution becomes
for
and
for
. That is, all of the electron states with
energy less than the Fermi energy are occupied, and all of the electron states with energy greater than the Fermi energy are unoccupied.
If you have an expression for the density of states
, the total free energy can be calculated as
problem, it turns out that the average energy per sodium atom is
.
. For the purposes of this
ANSWER:
= 1.94
Part B
What is the speed
of an electron that has energy
?
Express your answer in meters per second using three significant figures.
Hint 1. How to approach the problem
Assume that the energy of the electron is entirely in the form of kinetic energy. Use the standard formula for the kinetic energy to find the velocity of an electron in
sodium. It might help to convert
into joules.
Hint 2. Electron volts and joules
The conversion from electron volts to joules is
.
ANSWER:
= 8.250×105
Part C
At what Kelvin temperature
is
equal to
?
Express your answer in kelvins to three significant figures.
ANSWER:
= 3.740×104
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This temperature is called the Fermi temperature for the metal. It is approximately the temperature at which molecules in a classical ideal gas would have the same
kinetic energy as the fastest-moving electron in the metal.
± Understanding the Free-Electron Model of Metals
Description: ± Includes Math Remediation. Derive the equation for the Fermi energy, then calculate numeric values for gold.
Learning Goal:
To understand the Fermi-Dirac distribution and how it relates to the free-electron model of metals.
The free-electron model of metals assumes that the outermost valence electrons of each atom are free to roam about the metal. In its simplest form, the metal is modeled as a
box with zero potential energy inside and infinite potential energy outside. To describe the behavior of electrons in such a potential, we need to determine the allowed quantum
states and how the electrons are distributed among those states. The number of quantum states available with energies between and
is given by
, where
is the density of states; thus we have
,
where
divided by
is the number of states with energy less than or equal to the energy
.
,
is the volume of the box,
is the mass of an electron, and
is Planck's constant
Since electrons are fermions, they obey Pauli's exclusion principle, so you cannot have more than one electron in any state. The fraction of states that are filled at a particular
energy is given by the Fermi-Dirac distribution:
,
where
is Boltzmann's constant, is the absolute temperature, and
is an energy known as the Fermi energy. The Fermi-Dirac distribution is analogous to the MaxwellBoltzmann distribution from classical mechanics. It deals with indistinguishable particles that obey the exclusion principle instead of the classical, distinguishable particles of
the Maxwell-Boltzmann distribution. For metals, the Fermi energy is essentially constant, up to temperatures well above room temperature. Assume for the rest of the problem
that Fermi energy is a constant for any particular metal.
The total number of filled states (i.e., electrons in a particular state) with energy less than or equal to
states filled and then integrating over the energy from 0 to :
is found by multiplying the state density
and the fraction
of
.
To find the Fermi energy, consider the distribution of electrons among energy states at absolute zero (
). You would expect that at absolute zero there would be no
empty states at lower energy than a filled state (i.e., the electrons will be in the lowest state allowable by the exclusion prinicple).
Part A
What is the value of the Fermi-Dirac distribution for energies less than the Fermi energy, if the temperature is
?
Express your answer to two significant figures.
Hint 1. How to approach the problem
The Fermi-Dirac distribution has the form
.
You are asked to assume
, which makes the exponent negative. Notice that the absolute value of the exponent becomes infinite as
what happens to the exponential as the exponent grows toward negative infinity. Plug this limiting value into the Fermi-Dirac distribution.
. Consider
Hint 2. Value of the exponential
.
ANSWER:
= 1
Part B
What is the value of the Fermi-Dirac distribution for energies greater than the Fermi energy, if the temperature is
?
Express your answer to two significant figures.
Hint 1. How to approach the problem
The Fermi-Dirac distribution has the form
\large{\frac{1}{e^{(E-E_{\rm F})/(k_{\rm B}T)}+1}}.
You are asked to assume E>E_{\rm F}, which makes the exponent positive. Notice that the value of the exponent becomes infinite as T \rightarrow 0\;{\rm K}.
Consider what happens to the exponential as the exponent grows toward infinity. Plug this limiting value into the Fermi-Dirac distribution.
Value
of the exponential
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\lim_{x \rightarrow \infty} e^x = \infty.
ANSWER:
\texttip{f\left(E\right)}{f(E)} = 0
These results are as you would expect: All of the states below a certain energy are filled and none above that energy are filled. This provides useful information,
because this energy is the Fermi energy. Next, you will use this information to find an expression for the Fermi energy.
Part C
Using the values of \texttip{f\left(E\right)}{f(E)} that you found in the previous parts, find the number N(E_{\rm F}) of electrons at or below the Fermi energy by evaluating
the integral
\large{N(E_{\rm F})=\int_0^{E_{\rm F}} g(E)f(E)\,dE}.
Express your answer in terms of \texttip{E_{\rm F}}{E_F}, \texttip{m}{m}, \texttip{\hbar }{hbar}, \texttip{V}{V}, and \texttip{\pi }{pi}.
Hint 1. Simplifying the integral
Before trying to evaluate this integral, notice that in the interval [0,E_{\rm F}], f(E)=1. Now, substitute in the values of \texttip{f\left(E\right)}{f(E)} and \texttip{g\left(E
\right)}{g(E)} to obtain
\large{\int_0^{E_{\rm F}}\frac{(2m)^{3/2}V}{2\pi^2 \hbar^3}E^{1/2}\,dE}.
This integral looks daunting, but notice that the only piece that depends on \texttip{E}{E} is the E^{1/2} term. Thus, you can pull all of the other terms outside of the
integral to obtain the relatively simple expression
\large{\frac{(2m)^{3/2}V}{2\pi^2 \hbar^3} \left[ \int_0^{E_{\rm F}}E^{1/2}\,dE \right]}.
ANSWER:
N(E_{\rm F}) =
Part D
Still considering the T=0\;{\rm K} limit, what fraction of the total number \texttip{N_{\rm total}}{N_total} of free electrons in the metal will be at energies above the Fermi
energy?
ANSWER:
\texttip{N_{\rm total}}{N_total}
2N_{\rm total}/3
N_{\rm total}/2
N_{\rm total}/3
N_{\rm total}/4
0
Part E
At absolute zero, all of the free electrons in the metal have energies less than or equal to the Fermi energy, so N(E_{\rm F})=N_{\rm total}. Using this equality, you can
solve for the Fermi energy \texttip{E_{\rm F}}{E_F} and find
\large{E_{\rm F}=\frac{3^{2/3}\pi^{4/3}\hbar^2}{2m} \left( \frac{N_{\rm total}}{V} \right)^{2/3}}.
The term N_{\rm total}/V is called the free-electron density and is usually denoted \texttip{n}{n}. (Be sure not to confuse this number with the function \texttip{n\left(E\right)}
{n(E)}.) The free-electron density for gold is 5.90\times10^{28}\;/{\rm m^3}. What is the Fermi energy \texttip{E_{\rm Fgold}}{E_Fgold} of gold?
Express your answer in electron volts to three significant figures.
ANSWER:
\texttip{E_{\rm Fgold}}{E_Fgold} = 5.53
\rm eV
Part F
What is the speed \texttip{v_{\rm F}}{v_F} of an electron at the Fermi energy of gold? For now, neglect the effects of relativity.
Express your answer in meters per second to two significant figures.
ANSWER:
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\texttip{v_{\rm F}}{v_F} = 1.40×106
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\rm m/s
The parameter \texttip{v_{\rm F}}{v_F}, known as the Fermi speed, is one of the quantities that determines the resistivity of a substance.
Part G
What is the value of \texttip{\gamma }{gamma}, the relativistic correction factor, at this speed? Give your answer to six significant figures, even though none of the data
have this many significant figures.
Express your answer to six significant figures.
Hint 1. Formula for \texttip{\gamma }{gamma}
Recall that
\large{\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}},
where \texttip{c}{c} is the speed of light and \texttip{v}{v} is the speed of the object (technically the frame of reference).
ANSWER:
\texttip{\gamma }{gamma} = 1.00001
You were asked for the answer to six significant figures, because to the proper two significant figures \gamma=1. The tiny difference between \texttip{\gamma
}{gamma} and one shows that you were justified in neglecting relativistic effects.
Band Gap Energy Conceptual Questions
Description: Conceptual questions on the properties of a solid based on its energy band gap.
The energy bands of a hypothetical solid are illustrated in the figure.
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Part A
Is this solid a conductor, a semiconductor, or an insulator?
Hint 1. Distinguishing between conductors, semiconductors, and insulators
The energy band containing the highest-energy electrons in a solid is called the valence band. The next-highest band is called the conduction band. Distinguishing
between conductors, semiconductors, and insulators is based on the gap, or lack of gap, between these two bands.
In a conductor, the two bands touch or overlap. This means that for any available energy, no matter how small, a valence electron can absorb the energy and move
into an unoccupied state in the conduction band. This implies that when attached to a voltage source, electrons in a conductor are free to move because
unoccupied states are readily available.
In an insulator, the energy gap between the bands is quite large, typically several electron volts. This means that for an electron to move to an unoccupied state, it
would need to receive a few electron volts of energy. This amount of energy is typically not available to individual electrons, so electrons are not free to move in
response to an applied voltage.
In a semiconductor, there is an energy gap between the valence and conduction bands, but it is substantially smaller than in an insulator, typically around 1.0 \rm
{eV} or less. This amount of energy can sometimes be absorbed from either thermal motion or an applied voltage, so electrons are free to move to unoccupied
states and hence form a current, but not in the numbers typically present in a conductor.
ANSWER:
conductor
semiconductor
insulator
Imagine applying a relatively small potential difference across a sample of this material.
Part B
At temperatures near absolute zero, will any current flow?
Hint 1. Thermal energy at absolute zero
As a solid's temperature approaches absolute zero, the thermal energy available for exciting valence electrons approaches zero. Theoretically, at absolute zero, all
of the electrons in a semiconductor are "locked" into their lowest possible energy state. It is worth noting, however, that in conductors (metals) current can flow
even at zero temperature.
ANSWER:
yes
no
Part C
At 300 \rm K, will any current flow?
Hint 1. Thermal energy and temperature
The thermal energy available in a solid is directly proportional to the solid's temperature. Therefore, as the temperature is increased, more thermal energy is
available for exciting valence electrons into the conduction band.
ANSWER:
yes
no
Part D
As the temperature is increased above 300 \rm K, will more, less, or the same amount of current flow than at 300 \rm K?
Hint 1. Thermal energy and temperature
The thermal energy available in a solid is directly proportional to the solid's temperature. Therefore, as the temperature is increased, more thermal energy is
available for exciting valence electrons into the conduction band.
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ANSWER:
more
less
the same
Part E
If it were possible to increase the band gap energy to more than 0.7 \rm {eV}, would more, less, or the same amount of current flow than with the gap at 0.7 \rm {eV}?
Hint 1. Effect of band gap energy on current
In order for current to flow in a semiconductor, electrons must jump from the valence band to the conduction band. The probability of this jump occurring is strongly
dependent on the size of the band gap, with smaller gaps resulting in a substantially increased probability.
ANSWER:
more
less
the same
Exercise 42.26
Description: Pure germanium has a band gap of E_gap. The Fermi energy is in the middle of the gap. (a) For temperature of T_1 calculate the probability f(E) that a
state at the bottom of the conduction band is occupied. (b) For the temperature in part...
Pure germanium has a band gap of 0.67 {\rm {\rm eV}} . The Fermi energy is in the middle of the gap.
Part A
For temperature of 270 {\rm {\rm K}} calculate the probability f(E) that a state at the bottom of the conduction band is occupied.
ANSWER:
f(E) =
= 5.6×10−7
= 5.61×10−7,
Also accepted:
= 5.6×10−7,
= 5.56×10−7,
= 5.6×10−7
Part B
For the temperature in part \rm A, calculate the probability that a state at the top of the valence band is empty.
ANSWER:
f(E) =
= 5.6×10−7
= 5.61×10−7,
Also accepted:
= 5.6×10−7,
= 5.56×10−7,
= 5.6×10−7
Part C
For temperature of 300 {\rm {\rm K}} calculate the probability f(E) that a state at the bottom of the conduction band is occupied.
Express your answer using two significant figures.
ANSWER:
f(E) =
= 2.4×10−6
Also accepted:
= 2.37×10−6,
= 2.3×10−6,
= 2.35×10−6,
= 2.4×10−6
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For the temperature in part \rm C, calculate the probability that a state at the top of the valence band is empty.
Express your answer using two significant figures.
ANSWER:
f(E) =
= 2.4×10−6
= 2.37×10−6,
Also accepted:
= 2.3×10−6,
= 2.35×10−6,
= 2.4×10−6
Part E
For temperature of 340 {\rm {\rm K}} calculate the probability f(E) that a state at the bottom of the conduction band is occupied.
ANSWER:
f(E) =
= 1.1×10−5
= 1.09×10−5,
Also accepted:
= 1.1×10−5,
= 1.08×10−5,
= 1.1×10−5
Part F
For the temperature in part \rm E, calculate the probability that a state at the top of the valence band is empty.
ANSWER:
f(E) =
= 1.1×10−5
= 1.09×10−5,
Also accepted:
= 1.1×10−5,
= 1.08×10−5,
= 1.1×10−5
Exercise 42.27
Description: Germanium has a band gap of 0.67 eV. Doping with arsenic adds donor levels in the gap 0.01 eV below the bottom of the conduction band. At a
temperature of 300 K, the probability is 4.4 * 10^( - 4) that an electron state is occupied at the bottom of ...
Germanium has a band gap of 0.67 {\rm eV}. Doping with arsenic adds donor levels in the gap 0.01 {\rm eV} below the bottom of the conduction band. At a temperature of
300 {\rm K}, the probability is 4.4 \times 10^{ - 4} that an electron state is occupied at the bottom of the conduction band.
Part A
Where is the Fermi level relative to the conduction band in this case?
Express your answer using two significant figures.
ANSWER:
E_{\rm F} = 0.20
\rm eV below the conduction band.
Also accepted: 0.200, 0.20, 0.200, 0.20
Properties of Solids: The Energy Bands Model
Description: Brief review of energy bands model and relative (mostly) conceptual questions on insulators, semiconductors, and conductors.
To understand the physical and chemical properties of solids, it is important to become familiar with the concept of energy bands.
When a large number of atoms are brought together to form the crystal structure typical of solids, the outer (or valence) electrons in one atom become affected by the
electrical interactions with the other electrons in neighboring atoms. As a result, their wave functions begin to overlap and distort. Since the Pauli exclusion principle does not
allow two electrons with the same quantum configuration to be in the same energy state, the energy levels of the valence electrons shift slightly, and the energy levels become
closely spaced. These closely spaced energy levels can be thought of as a continuous distribution of energies within a band. Thus, each valence electron state, which is a
sharp energy level when the atom is far enough from other atoms that their interactions are negligible, has now “merged” to form an energy band.
Part A
The electrons in solids can be found ____________
ANSWER:
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in only certain discrete sharp energy states associated with their orbits.
in energy states that overlap so that more than one electron is associated with a given energy level.
in the same energy states as if the atoms forming the solid were far enough so that their interactions could be neglected.
in closely spaced energy levels that form a continuous distribution of energy within a certain range.
There are no allowed energy states in the gap between adjacent bands. The width of a gap is called the band gap, or energy gap, and determines the amount of energy that
an electron needs to jump to the next adjacent energy band.
In insulators and semiconductors, the highest energy band that is fully occupied is called the valence band, while the next-higher band is the conduction band. Since the
valence band is completely filled, to be able to move in response to an applied electric field, the electrons in the valence band have to jump to the conduction band, which
requires an amount of energy that is not ordinarily available. In a semiconductor the energy gap between the valence and conduction bands is considerably smaller than in
insulators.
In conductors, the highest energy band is only partially filled, and is often called the conduction band. Thus, electrons in the states near the top of the band have many
adjacent unoccupied states available, and only a small amount of energy is required to move an electron to one of these energy states.
Part B
When an electron in the valence band in an insulator gains additional energy, it can jump to ___________
ANSWER:
an adjacent energy state within the valence band.
an adjacent energy state in the conduction band.
an energy state in the conduction band only if it has gained an amount of energy at least equal to the band gap.
an energy state between the valence band and the conduction band.
Part C
When an electron in the highest energy band in a conductor gains additional energy, it can jump to ___________
ANSWER:
an adjacent energy state within the same band.
an energy state in the next-higher band.
an energy state just above the conduction band.
Part D
What is the main difference between insulators and semiconductors?
ANSWER:
In semiconductors, the valence band is only partially filled.
In semiconductors, the energy gap between the valence and conduction band is considerably smaller than in insulators.
At absolute zero, in semiconductors some electrons occupy energy states in the conduction band.
The main difference between an insulator and a semiconductor is the fact that in a semiconductor the energy gap between the valence and the conduction bands is
only on the order of 1 \rm eV, as opposed to 5 \rm eV for insulators. This explains why as the temperature of a semiconductor increases, more and more electrons
gain enough thermal energy to jump to the conduction band. Thus, the conductivity of semiconductors increases with temperature.
Part E
The energy gap between the valence band and the conduction band in germanium is only 0.67 \rm eV. If a germanium crystal is irradiated with electromagnetic waves of
8.80×10−7 {\rm m} wavelength, will some electrons in the top level of the valence band gain enough energy to jump to the lowest energy level in the conduction band?
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Hint 1. Energy of electromagnetic waves
The energy of an electromagnetic wave is
\large{E=\frac{hc}{\lambda}},
where \texttip{\lambda }{lambda} is the wavelength of the wave, \texttip{c}{c} is the speed of light, and \texttip{h}{h} is Planck's constant.
Hint 2. Energy gap
Recall that the energy gap between the valence band and the conduction band is equal to the amount of energy that an electron needs to jump from the valence
band to the conduction band.
ANSWER:
yes
no
Exercise 42.28
Description: (a) Suppose a piece of very pure germanium is to be used as a light detector by observing, through the absorption of photons, the increase in conductivity
resulting from generation of electron-hole pairs. If each pair requires ## eV of energy, what...
Part A
Suppose a piece of very pure germanium is to be used as a light detector by observing, through the absorption of photons, the increase in conductivity resulting from
generation of electron-hole pairs. If each pair requires 0.71 {\rm \;eV} of energy, what is the maximum wavelength that can be detected?
Express your answer with the appropriate units.
ANSWER:
\lambda =
= 1.7×10−6
= 1.75×10−6
Also accepted:
,
= 1.7×10−6
Part B
In what portion of the spectrum does it lie?
ANSWER:
ultraviolet
infrared
visible light
gamma rays
Part C
If the material is silicon, with an energy requirement of 1.20 {\rm \;eV} per pair, corresponding to the gap between valence and conduction bands in that element, what is
the maximum wavelength that can be detected?
Express your answer with the appropriate units.
ANSWER:
\lambda =
= 1.0×10−6
Also accepted:
= 1.03×10−6
,
= 1.0×10−6
Part D
In what portion of the spectrum does it lie?
ANSWER:
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gamma rays
visible light
ultraviolet
infrared
Exercise 42.29
Description: At a temperature of 290 K, a certain p-n junction has a saturation current I_S = 0.500 mA. (a) Find the current at this temperature when the voltage is 1.00
mV, -1.00 mV, 100 mV, and -100 mV. (b) Is there a region of applied voltage...
At a temperature of 290 {\rm K}, a certain p-n junction has a saturation current I_{\rm S} = 0.500 {\rm mA}.
Part A
Find the current at this temperature when the voltage is 1.00 {\rm mV}, -1.00 {\rm mV}, 100 {\rm mV}, and -100 {\rm mV}.
Enter your answers numerically separated by commas. Express your answers using two significant figures.
ANSWER:
I_{1.00}, I_{-1.00}, I_{100}, I_{-100} = 2.0×10−2, −2.0×10−2, 27, -0.49
\rm mA
Also accepted: 2.04×10−2, −1.96×10−2, 26.8, -0.491, 2.0×10−2, −2.0×10−2, 27, -0.49, 2.04×10−2, −1.96×10−2, 26.9, -0.491, 2.0×10−2, −2.0×10−2, 27, -0.49
Part B
Is there a region of applied voltage where the diode obeys Ohm's law?
ANSWER:
yes
no
The Basics of Nuclear Physics
Description: Standards for nuclear nomenclature are given and then used in basic problems.
Learning Goal:
To understand the notation and basic quantities involved in nuclear physics.
Nuclear physics borrows the symbols for elements from chemistry. However, knowing which element we are dealing with only tells us one of the numbers important to nuclear
physics. When referring to a specific nucleus, we use the following notation:
^{56} _{26} {\rm Fe}.
The superscript, 56 in this case, denotes the total number of nucleons (protons and neutrons) in the nucleus. This is called the nucleon number and is given the symbol
\texttip{A}{A}. The subscript, 26 in this case, is the number of protons in the nucleus. This is called the atomic number and is given the symbol \texttip{Z}{Z}. An atom's atomic
number determines which element the atom is, in this case {\rm Fe} (iron). Another important number characterizing an atom is the neutron number \texttip{N}{N}. Since
\texttip{A}{A} is the total number of nucleons in the nucleus, the neutron number may be found from the equation A=Z+N. Solving for \texttip{N}{N} gives N=A-Z. Nuclei with the
same atomic number but different neutron numbers are called isotopes. Isotopes are often written in a form such as "iron-56."
Part A
What is the atomic number \texttip{Z}{Z} of ^{7} _{3} {\rm Li}?
Express your answer as an integer.
ANSWER:
\texttip{Z}{Z} = 3 protons
Part B
What is the nucleon \texttip{A}{A} number of carbon-14?
Express your answer as an integer.
ANSWER:
\texttip{A}{A} = 14 nucleons
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What is the neutron number \texttip{N}{N} of ^{56} _{26} {\rm Fe}?
Express your answer as an integer.
ANSWER:
\texttip{N}{N} = 30 neutrons
Part D
Which of the following choices lists a pair of isotopes?
ANSWER:
^{14} _{\ 6} {\rm C} and ^{14} _{\ 7} {\rm N}
^{3} _{2} {\rm He} and ^{6} _{3} {\rm Li}
^{15} _{\ 7} {\rm N} and ^{16} _{\ 8} {\rm O}
^{3} _{1} {\rm H} and ^{2} _{1} {\rm H}
It has been found that the radii of most nuclei are well approximated by the equation \large{R=R_0A^{\frac{1}{3}}}, where R_0=1.2 \times 10^{-15}\; {\rm m} is an
experimentally determined constant. Because the mass of a nucleon (a proton or a neutron) is close to one atomic mass unit (1 \;{\rm u}), the nucleon number \texttip{A}{A} is
sometimes called the mass number.
Part E
What is the approximate radius \texttip{R}{R} of ^{208} _{\ 82} {\rm Pb}?
Express your answer in meters to two significant figures.
Hint 1. What is \texttip{A}{A}?
Which of the following numbers correctly gives the value of \texttip{A}{A} for ^{208} _{\ 82} {\rm Pb}?
ANSWER:
290
208
126
82
ANSWER:
\texttip{R}{R} = 7.10×10−15
{\rm m}
Part F
Assuming that each nucleus is roughly spherical and that its mass is roughly equal to \texttip{A}{A} (in atomic mass units {\rm u}), what is the density \texttip{\rho }{rho} of
a nucleus with nucleon number \texttip{A}{A}?
Express your answer in terms of \texttip{A}{A}, \texttip{R_{\rm 0}}{R_0}, and \texttip{\pi }{pi}.
Hint 1. Find the volume of the nucleus
Recall that the volume of a sphere is given by V= (4/3) \pi R^3. Assuming that the nucleus is spherical, find its volume.
Express your answer in terms of \texttip{A}{A}, \texttip{R_{\rm 0}}{R_0}, and \texttip{\pi }{pi}.
ANSWER:
\texttip{V}{V} =
ANSWER:
\texttip{\rho }{rho} =
{\rm u/m^3}
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The fact that \texttip{A}{A} does not appear in your result is a very important property of nuclei. The observation that most nuclei have about the same density was
key to the development of George Gamow's liquid drop model of nuclear binding.
NMR Spectroscopy
Description: Find the magnetic resonance frequency for a proton in a given magnetic field. Calculate the changes in frequency due to chemical shifts.
Chemists use a wide array of techniques for determining the exact composition and structure of a compound. One of the most robust and interesting of these is nuclear
magnetic resonance (NMR) spectroscopy. In this problem, you will see how NMR spectroscopy allows for the precise determination of the structure of an organic compound.
Nuclei with odd numbers of neutrons or protons have a magnetic moment. In the presence of a strong magnetic field, some nuclei will align parallel and some will align
antiparallel to the field. If a sample is subjected to electromagnetic radiation with photon energy equal to the difference in energy between the two nuclear alignment states,
some nuclei in parallel states will absorb a photon and flip to antiparallel states.
Part A
If protons with magnetic moment \texttip{\mu }{mu} in the z direction are in a strong magnetic field of magnitude \texttip{B}{B} in the z direction, what is the frequency
\texttip{f}{f} of radiation that will be absorbed by the proton as it transitions from parallel to antiparallel states?
Express your answer in terms of \texttip{\mu }{mu}, \texttip{B}{B}, and Planck's constant \texttip{h}{h}.
Hint 1. Find the difference in energy between the two states
What is the (positive) difference in energy between the antiparallel and parallel states? Recall that the potential energy due to magnetic moment is given by
U=-\vec{\mu}\cdot \vec{B}.
Express your answer in terms of \texttip{\mu }{mu} and \texttip{B}{B}.
ANSWER:
\texttip{\Delta E}{Delta E} =
Hint 2. Relating energy and frequency
For a photon, E=hf.
ANSWER:
\texttip{f}{f} =
Part B
Suppose that a small shift \texttip{\Delta f}{Deltaf} in the frequency of the absorbed radiation is observed. This shift is attributed to a small change \Delta \mu in the
magnetic moment. What is the value of \Delta \mu?
Express your answer in terms of \texttip{\Delta f}{Deltaf}, \texttip{B}{B}, and \texttip{h}{h}.
Hint 1. How to approach the problem
If there are small shifts in the frequency ( \texttip{\Delta f}{Deltaf}) and magnetic moment (\Delta \mu), then the frequency and magnetic moment will be f+\Delta f
and \mu+\Delta \mu, respectively. Plug these values into the equation that you derived in Part A. Now, solve for \Delta \mu. If you have variables in your solution
that are not asked for in the answer instructions, you should be able to substitute expressions with the correct variables to eliminate the unwanted variables.
ANSWER:
\Delta \mu =
In practice, small frequency shifts of this type are caused by the differing density of the electron cloud around protons in an organic molecule. The density of the
electron cloud is determined by the atoms near the proton. Since oxygen holds electrons more strongly than carbon, there will be a small increase in the magnetic
moment for protons near oxygen atoms, because the electrons screen the proton less than they would if there were only carbon atoms nearby. A number of bond
types and functional groups can be discerned by the small frequency shifts, known as chemical shifts, that they induce in the NMR absorption spectrum.
Part C
The magnetic moment of a hydrogen nucleus is roughly 2.82\times 10^{-26}\;\rm J/T. What would be the resonant frequency \texttip{f}{f} in a 5.00 \rm T magnetic field?
Express your answer in hertz to three significant figures.
ANSWER:
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\texttip{f}{f} = 4.250×108
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\rm Hz
Part D
A hydrogen atom bonded directly to an oxygen atom has a shift in resonant frequency of 12 parts per million. What is the change in the hydrogen atom's magnetic
moment \Delta \mu caused by the presence of a nearby oxygen atom in a 5.00 \rm T magnetic field?
Express your answer in joules per tesla to two significant figures.
Hint 1. How to approach the problem
Find the absolute change in frequency from the relative (parts per million) change. Then, use the equation you derived in Part B to find the change in magnetic
moment.
Hint 2. The meaning of parts per million
Parts per million can more easily be understood if you think of percent as "parts per hundred." ("Cent" is from the latin word for one hundred, so percent literally
means "per hundred.") Like percent, parts per million is a measure of relative change. A 1\% change in the frequency would mean \Delta f/f =1/100. Similarly, a
change of one part per million would mean \Delta f/f =1/1,000,000
ANSWER:
\Delta \mu = 3.40×10−31
\rm J/T
Other important pieces of information can be gleaned from an NMR spectrum. The ratio of hydrogen atoms in particular functional groups can be determined from the
ratio of the areas underneath the peaks that correspond to the chemical shifts for those particular functional groups. Also, nearby protons can cause peak splitting,
allowing you to determine the number of nearby hydrogen atoms. All of this information can be used to get very detailed pictures of the structure of an unknown
molecule. In addition, many atoms besides hydrogen have similarly useful NMR spectra.
Exercise 43.6
Description: The most common isotope of uranium, ^238_92U, has atomic mass 238.050788 u. (a) Calculate the mass defect. (b) Calculate the binding energy. (c)
Calculate the binding energy per nucleon.
The most common isotope of uranium, ^{238}_{92}{\rm U}, has atomic mass 238.050788 {\rm u}.
Part A
Calculate the mass defect.
Express your answer using three significant figures.
ANSWER:
\Delta m = 1.93
{\rm u}
Part B
Calculate the binding energy.
Express your answer using three significant figures.
ANSWER:
E = 1800
{\rm MeV}
Part C
Calculate the binding energy per nucleon.
Express your answer using three significant figures.
ANSWER:
E_0 = 7.57
{\rm MeV} per nucleon
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