AP Atomic and Nuclear Review

Report
Do Now (3/24/14):
• What are the major topics and concepts that
we covered in our atomic and nuclear units?
AP Atomic and Nuclear Review
Jessalyn Timson
3/24/14
Nuclear Physics - Multiple Choice
Practice Questions for AP Physics B
“Maturity is often more absurd than
youth and very frequently is most
unjust to youth”
– Thomas A. Edison
(1) The density of nuclear matter is
(a) directly proportional to the number protons in the
nucleus
(b) directly proportional to the number neutrons in the
nucleus
(c) directly proportional to the square of the number of
nucleons in the nucleus
(d) directly proportional to the 4th power of the number
of nucleons in the nucleus
(e) independent of the number nucleons in the nucleus
• The correct option is (e). Nuclear radius R is given
by
• R =R0A1/3
• where R0 = 1.2×10–15 m and A is the mass number
(or, nucleon number).
• The volume of the nucleus which is proportional
to R3 is therefore proportional to the mass
number A. Since the density is the ratio of mass
to volume, it follows that the density of the
nucleus is independent of the mass number and
is constant (approximately 2.3×1017 kgm–3).
(2) Nuclear force is
(a) short range and charge dependent
(b) long range and charge dependent
(c) short range and charge independent
(d) long range and charge independent
(e) electromagnetic in nature
• Nuclear force is a short range force produced by the
exchange of π-mesons between the nucleons.
• The force between a proton and a neutron is produced
because of the exchange of charged π-mesons (π+ and
π–) where as the force between two protons and that
between two neutrons is produced by the exchange of
uncharged π-mesons (π0).
• Nuclear force is a strong attractive force (in fact, the
strongest natural force) and is charge independent.
• The correct option is (c).
(3) In the nuclear reaction given by
14 + He4 = Xn + H1, what is the nucleus Xn?
N
7
2
m
1
m
(a) Oxygen of mass number 18
(b) Nitrogen of mass number 18
(c) Oxygen of mass number 16
(d) Oxygen of mass number 17
(e) Nitrogen of mass number 17
• Since the total mass number on the left hand side
of the equation is 18, the mass number of X has
to be 17 (for balancing the equation).
• [14 + 4 = n + 1 from which n = 17]
• The atomic number of X (or, the number of
protons in the nucleus X) is 8 since the total
atomic number on the left hand side of the
equation is 9.
• [7 + 2 = m + 1 from which m = 8]
• Therefore the nucleus X is that of oxygen (8O17)
(4) Cobalt 60 is a radioactive source with a half life
of 5.27 years. You may take it as 5 years. After
how many years will the activity of a sample of
cobalt 60 be decreased to 1/16 its original
activity?
(a) 10 years
(b) 16 years
(c) 20 years
(d) 32 years
(e) 40 years
• The activity (number of disintegrations per second) is
directly proportional to the number of undecayed nuclei in
the sample. Since the number of undecayed nuclei reduces
to half the initial value in every half life period, the activity
also reduces to half the initial value in every half life period.
Therefore, after 5 years the activity becomes half the initial
value; after 10 years the activity becomes 1/4 the initial
value; after 15 years it becomes 1/8 the initial value and
after 20 years the activity becomes 1/16 the initial value.
The correct option is (c).
• [The activity A after n half lives is given by A = A0/2n where
A0 is the initial activity. Therefore, 1/16 = 1/2n so that 2n =
16 from which n = 4. Four half lives = 20 years].
(5) In a typical fission reaction, a U235 nucleus absorbs a
slow neutron and becomes a compound nucleus U236 in
a highly excited state. U236 then undergoes fission,
producing two fission fragments (Xe140 and Sr94) and
two neutrons. Typically what should be the energy of
the slow neutron that initiates the fission reaction in
U235 nucleus?
(a) 25 Mev
(b) 2.5 KeV
(c) 250 eV
(d) 25 eV
(e) 0.025 eV
• Neutrons can be absorbed by the U235 nuclei if
they are of thermal energies. Such thermal
neutrons have energies of the order of a small
fraction of an electron volt as given in option
(e).
What is the atomic number of
236 U?
92
A. 92
b. 144
c. 236
d. 328
What is the atomic mass number
of 23692U?
A. 92
b. 144
c. 236
d. 328
What is the neutron number of
236 U?
92
A. 92
b. 144
c. 236
d. 328
What is the binding energy of the
12 C nucleus?
6
A. 12 MeV
b. 937 keV
c. 92 MeV
210
84Po
decays by alpha decay. The
resulting isotope is
a. 21082Pb
b. 21081Bi
c. 20881Bi
206
d. 82Pb
What is the energy released in the
above decay?
210
( 84Po decays by alpha decay.)
a. 234 MeV
b. 23.4 MeV
c. 5.4 MeV
d. 13.6 eV
Who formulated the "Uncertainty
Principle”?
A. Niels Bohr
b. Max Born
c. Werner Heisenberg
d. Wolfgang Pauli
e. Erwin Schrodinger
Who first found the wave
equation?
A. Niels Bohr
b. Max Born
c. Werner Heisenberg
d. Wolfgang Pauli
e. Erwin Schrodinger
What is the energy of the n = 3
quantum state of the hydrogen
atom?
A. -9 eV
b. -4.5 eV
c. -1.5 eV
d. -.5 eV
What is the energy of the electron
in the ground state of doubly
ionized lithium, Li2+?
A. -13.6 eV
b. -40.8 eV
c. -100 eV
d. -122.4 eV
Estimate the energy of an electron
in the ground state of Pb81+ lead
with only 1 electron.
A. -13.6 KeV
b. -91 KeV
c. -1.1 KeV
d. -13.6 eV
In a nuclear reactor, uranium fissions into krypton and
barium via the reaction
What are the mass number A and atomic number Z of
the resulting krypton nucleus?
A—The total number of protons + neutrons is
conserved. Before the reaction, we have one free
neutron plus 235 protons and neutrons in the
uranium, for a total of 236 amu. After the reaction,
we have 141 amu in the barium plus 3 free
neutrons for a total of 144 amu… leaving 92 AMUs
for the krypton.
Charge is also conserved. Before the reaction, we
have a total charge of +92 from the protons in the
uranium. After the reaction, we have 56 protons in
the barium. Since a neutron carries no charge, the
krypton must account for the remaining 36
protons.
In a nuclear reactor, uranium fissions into
krypton and barium via the reaction
How much mass is converted into the kinetic
energy of the resulting nuclei? 1 amu
2 amu
3 amu
zero
much less than 1 amu
E—Einstein's famous equation is written
ΔE = mc2, because it is only the lost mass
that is converted into energy. Since we
still have a total of 236 amu after the
reaction, an entire amu of mass was not
converted to energy. Still, the daughter
particles have kinetic energy because
they move. That energy came from a
very small mass difference, probably
about a million times less than one amu.
decays via β+ emission. Which of
the following is the resulting
nucleus?
C—In β+ emission, a positron is ejected
from the nucleus. This does not change
the total number of protons + neutrons,
so the atomic mass A is still 15.
However, charge must be conserved.
The original O nucleus had a charge of
+8, so the products of the decay must
combine to a charge of +8. These
products are a positron, charge +1, and
the daughter nucleus, which must have
charge +7.
Compared to Hydrogen, 11H, the
element Helium, 42He, has
a) more mass and is larger in size
b) more mass and is about the same
in size
c) more mass and is smaller in size
d) none of the above
c - more mass but a smaller radius
Solution: The nucleus of helium has four
nucleons compared to hydrogen's one, so it
is about four times as massive as hydrogen.
The nucleus of helium has twice the electric
charge of hydrogen, and pulls its electrons
into a tighter orbit than does hydrogen to its
solitary electron. Therefore, helium is
smaller, in size than the hydrogen atom.
Both nuclear fission and nuclear fusion release enormous
energy. When a uranium nucleus fissions, the released energy
is mainly kinetic energy of the repelling fragments. When a
pair of hydrogen isotopes fuse, the energy initially released is
in the form of
a) gamma radiation.
b) kinetic energy of recoiling particles.
c) potential energy of the helium nucleus that is formed.
d) heat.
e) a combination of all of the above
b - kinetic energy of the recoiling particles
Solution: The energy initially and typically released in the fusion
of hydrogen isotopes is divided between the kinetic energy of the
two particles produced a helium nucleus and a neutron.
Interestingly, a pair of hydrogen isotopes can't fuse to produce a
lone helium nucleus - even though the numbers of protons and
neutrons add up correctly.
Why? Momentum and energy conservation: If a lone helium
nucleus flies away after the reaction, it adds momentum that
wasn't there before. Or if it remains motionless, there's no
mechanism for energy release. So it can't move and it can't sit
still!
A fusion reaction requires the creation of at least two particles to
share the released energy - or, in some cases, like in the dense
core of the sun, a neighboring nucleus to take up some of the
energy.
When a U-235 nucleus absorbs a neutron
and undergoes nuclear fission, about 200
MeV of energy is released. But in what
form?
Interestingly, most of this energy initially
appears in the form of
a) gamma rays
b) KE of emitted neutrons
c) KE of the fission fragments
d) heat
e) each of these, about equally
c - KE of the fission fragments
Solution: Some energy is emitted in the
form of gamma rays and some goes into
the kinetic energy of emitted neutrons,
but most of the energy of nuclear fission is
in the kinetic energy of the fission
fragments. The positively-charged
fragments repel each other and fly apart
at high speed. Soon their energy is shared
among many atoms as internal energy. It
then spreads as heat
When a neutron interacts with a U235 nucleus, it can fission many
possible ways. What element results if
it fissions into two identical nuclei?
If uranium fissions into two
identical elements, their atomic
number is half 92, or 46. That's
palladium.
How many neutrons are produced
when a U-235 nucleus fissions into
Sr-90 and Xe-138?
8 neutrons
Solution: If U-235 fissions into
strontium-90 and xenon-138, 8 neutrons
are released, according to the reaction
shown below. The number of neutrons
released per fission reaction for most
reactions is considerably less than 8!
Some common fusion reactions of hydrogen
isotopes are shown in incomplete forms. Can
you complete them?
Some common fusion reactions of hydrogen
isotopes are shown in incomplete forms. Can
you complete them?
Some common fusion reactions of hydrogen
isotopes are shown in incomplete forms. Can
you complete them?
E. The remaining material is
manganese.
Solution: Beta emission is the emission
of an electron, which effectively
transforms a neutron to a proton,
increasing the atomic number of the
radioactive element by 1. A look at the
periodic table shows that manganese,
atomic number 25, is 1 greater than
chromium, atomic number 24.
I. 7.5 grams
Solution: After 14 minutes, 4 halflives of 3.5 minutes each have
elapsed (14/3.5 = 4). The 120
grams is therefore halved 4 times
to leave 7.25 grams.
A hypothetical atom has two energy
levels, as shown above. (a) What
wavelengths of electromagnetic
radiation can be absorbed by this atom?
Indicate which of these wavelengths, if
any, represents visible light.
ΔE = hc/λ, so hc/ΔE = λ. hc = 1240 eV·nm, as
found on the equation sheet. ΔE represents
the difference in energy levels. An electron in
the ground state can make either of two
jumps: it could absorb a 2.1-eV photon to get
to the middle energy level, or it could absorb
a 3.3 eV photon to escape the atom. An
electron in the middle state could absorb a
1.2-eV photon to escape the atom. That
makes three different energies.
Convert these energies to wavelengths using
ΔE = hc/λ, so hc/ΔE = Δ. hc = 1240 eV·nm, as
found on the equation sheet; ΔE represents
the energy of the absorbed photon, listed
above. These photons thus have wavelengths
of 590 nm for the E1 to E2 transition; 380 nm
or less for the E1 to E∞ transition; and 1030
nm or less for the E2 to E∞ transition. Only
the 590 nm wavelength is visible because the
visible spectrum is from about 400–700 nm.
A hypothetical atom has two energy levels,
as shown above.
(b)Now, monochromatic 180-nm ultraviolet
radiation is incident on the atom, ejecting an
electron from the ground state. What will be:
the ejected electron's kinetic energy
Find the energy of the incident
photon using ΔE = hc/λ, getting 6.9
eV. This is the total energy absorbed
by the electron, but 3.3 eV of this is
used to escape the atom. The
remaining 3.6 eV is kinetic energy.
A hypothetical atom has two energy levels,
as shown above.
(b)Now, monochromatic 180-nm ultraviolet
radiation is incident on the atom, ejecting an
electron from the ground state. What will be
the ejected electron's speed:
To find speed, set kinetic energy equal to
1/ mv2. However, to use this formula, the
2
kinetic energy must be in standard units of
joules. Convert 3.6 eV to joules by
multiplying by 1.6 × 10-19 J/eV (this
conversion is on the constant sheet),
getting a kinetic energy of 5.8 × 10-19 J.
Now solve the kinetic energy equation for
velocity, getting a speed of 1.1 × 106 m/s.
This is reasonable—fast, but not faster
than the speed of light.
A hypothetical atom has two energy levels,
as shown above.
(b)Now, monochromatic 180-nm ultraviolet
radiation is incident on the atom, ejecting an
electron from the ground state. What will be:
the incident photon's speed
A photon is a particle of light.
Unless it is in an optically dense
material (which the photons here
are not), the speed of a photon is
always 3.0 × 108 m/s.
A hypothetical atom has two energy levels, as shown
above. For parts (c) and (d), imagine that the 180-nm
radiation ejected an electron that, instead of being in
the ground state, was initially in the –1.2 eV state.
(c) Would the speed of the ejected electron increase,
decrease, or stay the same? Justify your answer briefly.
(c) The electron absorbs the same
amount of energy from the incident
photons. However, now it only takes
1.2 eV to get out of the atom, leaving
5.7 eV for kinetic energy. With a larger
kinetic energy, the ejected electron's
speed is greater than before.
A hypothetical atom has two energy levels, as
shown above. For parts (c) and (d), imagine that
the 180-nm radiation ejected an electron that,
instead of being in the ground state, was initially
in the –1.2 eV state.
(d) Would the speed of the incident photon
increase, decrease, or stay the same? Justify your
answer briefly.
(d) The speed of the photon is still
the speed of light, 3.0 × 108 m/s.
Is this decay possible?
And if so, would this reaction require
energy, or go by itself and yield energy?
yes, but it would require the input of
energy to make it start
Solution: The decay is possible for
both charge and mass number are
conserved. The numbers on the left
equal the numbers on the right.
Inspection of the mass/nucleon vs. atomic number
curve shown below reveal that the reaction
"climbs" the hill, so there is more mass after the
reaction than before.
This means that +the reaction will require an input
of energy. How much? An amount equal to the gain
in mass multiplied by the speed of light squared!
Suppose you could add or subtract
protons from oxygen nuclei to turn
oxygen into a gas that would glow
red when an electric current flows
through it. Would you add or
subtract protons? How many?
Add two protons to each nucleus
of oxygen and you increase the
atomic number from 8 to 10. You
then have neon, which will glow a
very nice red when a current flows
through it.

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