Homework Assignment 7 Physics 55 Problem 1: Problems from the Text Made available:

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Homework Assignment 7
Physics 55
Made available:
Due in class:
Tuesday, November 1, 2005
Monday, November 7, 2005
Problem 1: Problems from the Text
1. Do problem 21 on page 519 of the text.
2. Do problem 21 on page 541.
3. Do problem 24 on page 542 of the text.
4. Do problem 29 on page 542 of the text.
Problem 2: Some Problems Not in the Textbook
1. Some practice with nuclear reactions.
(a) The radioactive phosphorus nucleus
trino ν̄
32
15 P
beta decays by emitting an electron e and an antineu-
32
15 P
0
→ xy X + −1
e + 00 ν̄,
(1)
where ν is the lower-case Greek letter nu. An anti-neutrino is a neutral particle of negligible mass
but that can have a kinetic energy.
i. What is the atomic number y, the atomic mass number x, and the element X that results
from this decay? (You will likely need access to a periodic table of elements to determine the
element X.)
ii. If the maximum energy of the emitted electron is 1.7 MeV, what is the atomic mass of the
daughter nucleus X in atomic mass units u?
Data and a hint: An MeV (million electron volts) is a widely used unit of energy for nuclear
and particle reactions, with 1 MeV ≈ 1.6 × 10−13 J. An atomic mass unit u is a widely used
unit of mass for nuclei, with 1 u ≈ 1.66 × 10−27 kg, about the mass of one proton. From a
table of isotopes, one can look up that the mass of a 32
15 P nucleus is 31.973907 u. Finally, use
conservation of mass-energy to deduce the mass of the daughter nucleus X: total mass-energy
before the reaction (simply the rest energy mc2 of the phosphorus nucleus) has to equal the
total mass-energy after (the rest energy of mystery element X plus the kinetic energy of the
electron). For this problem, you can assume that the nucleus X has zero speed after the decay
(it is so massive compared to an electron that its speed is negligible) and you can ignore the
kinetic energy of the antineutrino, which is zero when the emitted electron has its maximum
kinetic energy.
(b) As we will see in Chapter 17 of the text, in a later stage of evolution, some stars begin to fuse
two 126 C nuclei (each of mass 12.000000 u) into one 24
12 Mg nucleus (which has a mass of 23.985042 u).
i. In MeV energy units, how much energy is released by this fusion reaction?
ii. You can figure out what the minimum absolute temperature T the core of the star must have
for this carbon fusion reaction to occur by calculating the minimum speed two +6-charged
carbon nuclei must have to approach each other such that the nuclei are able to overcome their
enormous electrical repulsion and just touch each other, in which case the short-ranged strong
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interaction becomes involved and a nuclear reaction can occur. This speed then corresponds
to a certain large temperature.
The first step is for you to calculate the electrical (also called Coulombic) potential energy of
two carbon nuclei in MeV units when the two nuclei are just touching each other. (This is the
amount of energy that will be released when the two nuclei are allowed to move away from
each other a large distance.) You can calculate the electrical potential energy by knowing
that the radius of a carbon nucleus is about 3 × 10−15 m, that the charge Q = 6e of a carbon
nucleus is six times the charge e of a proton (with value e ≈ 1.6 × 10−19 C, where the letter C
denotes a coulomb of charge), and that the formula for the electrical potential energy of two
particles with charges Q1 and Q2 that are a distance d apart is given by the equation
Eelectrical = 9.0 × 109
Q1 Q2
,
d
(2)
where the charges Q1 and Q2 are measured in units of coulombs, the distance d is measured
in units of meters m, and the energy Eelectrical has units of joules J.
iii. Each carbon nucleus needs to have an initial kinetic energy that is at least half of the above
electrical potential energy, if they are to overcome the electrical repulsion and approach within
a radius of each other. The kinetic theory of gases relates the average kinetic energy K of a
molecule in a gas of temperature T by the formula
K=
3
kT,
2
(3)
where the the letter k ≈ 1.4×10−23 J/K is the so-called Boltzmann constant. Use this equation
to deduce the minimum temperature that the core of a star must have for carbon-carbon
fusion reactions to occur, giving your answer as a multiple of the Sun’s core temperature of
15,000,000 K.
2. (a) Assuming that the luminosity and the rate of hydrogen fusion into helium in the Sun remains
constant throughout the lifetime of the Sun, determine what fraction of the Sun’s mass will be
converted into helium over the remaining 5 billion years of the Sun’s life.
(b) About how many Earths does this fraction of the Sun’s mass correspond to? (When writing
science articles for the general public, authors like to relate some big number to some image that
is easily appreciated by the reader.)
(c) Given that the current percentage composition (by mass) of the Sun is 70% hydrogen, 28% helium,
and 2% heavier elements, determine the composition of the Sun after 5 billion more years of fusion.
3. The star Sirius has a luminosity that is 23.5 times the luminosity of the Sun and has a mass that is
2.3 times the mass of the Sun. Explain whether the lifetime of Sirius will be longer, shorter, or about
the same as the lifetime of the Sun.
4. The visual binary 70 Ophiuchi has a period of 87.7 years. The parallax angle of 70 Ophiuchi is 0.2”
and the apparent length of the semimajor axis as seen through a telescope is 4.5”.
(a) What is the distance from Earth to 70 Ophiuchi in light years?
(b) What is the length of the semimajor axis in AU?
(c) What is the sum of the masses of the two stars, in units of the mass of the Sun?
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Problem 3: Extra Credit Problems
1. Use conservation of mass-energy and conservation of momentum to show that a proton p must have
a kinetic energy of at least 3.23 MeV for the nuclear reaction 136 C(p, n)137 N to occur. Without taking
conservation of momentum into account and just using conservation of mass-energy, one would conclude
incorrectly that a smaller minimum kinetic energy of 3.00 MeV would suffice for this reaction to occur.
Some data: the mass of a hydrogen atom is 1.007825 u, of the 136 C atom is 13.003355 u, of the nitrogen
atom 137 N is 13.005739 u, and of the neutron is 1.08665. A similar issue arises in chemical reactions,
a greater initial kinetic energy (called the threshold energy) is needed for an incoming atom than the
amount based just on energy conservation for some reaction to occur because of the need to conserve
momentum.
2. In lecture, I showed a picture of a bubble chamber in which an invisible photon (invisible since it was
uncharged and so did not create a trail of bubbles) passed near some nucleus and then disappeared,
producing an electron-positron pair in its place. Explain why a photon of arbitrarily high energy in
vacuum can not simply convert into an electron-positron pair, the photon has to pass near some third
massive particle like a nucleus.
Hint: Think about this problem from a reference frame moving with the center of mass of the electronpositron pair after they have been created and deduce a contradiction.
3. Explain why it is impossible for an isolated free electron (actually, for any point particle with no
internal structure) to absorb or emit a photon.
Problem 4: Comments about the Homework and Course
• About how long did it take you to complete this assignment?
• Do you feel that you are understanding the course material? If not, please indicate what topics or
ideas you would like to understand better.
• Comments or suggestions about other parts of the course such as reading, homeworks, lectures, or
observation sessions?
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