BROCK UNIVERSITY
Final Exam: April 2015
Course: PHYS 1P22/1P92
Examination date: 22 April 2015
Time of Examination: 9:00–12:00
Number of
Number of
Number of
Instructor:
pages: 12 (+ formula sheet)
students: 95
hours: 3
S. D’Agostino
A formula sheet is attached at the end of the test paper. No other aids are permitted except
for a non-programmable, non-graphing calculator. Use or possession of unauthorized materials
will automatically result in the award of a zero grade for this examination.
Solve all problems in the space provided.
Total number of marks: 50
SOLUTIONS
1. [5 marks] An electron is released from rest at the negative plate of a parallel plate
capacitor and accelerates towards the positive plate. The plates are separated by a distance
of 1.2 cm, and the electric field within the capacitor has a magnitude of 2.1 × 106 V/m.
What is the speed of the electron just as it reaches the positive plate?
Solution: The potential difference between the two plates is
∆V = Ed
∆V = 2.1 × 106 1.2 × 10−2
∆V = 2.52 × 104 V
When the electron moves between the two plates, the magnitude of the electric potential
energy that the electron loses, which is equal to the kinetic energy the electron gains, is
1 2
mv = |q||∆V |
2
2|q||∆V |
v2 =
r m
2|q||∆V |
v=
m
r
2 (1.6 × 10−19 ) (2.52 × 104 )
v=
9.11 × 10−31
v = 9.4 × 107 m/s
This is about 30% of the speed of light, so all you physics majors might wish to repeat
the calculation using special relativity to see whether there is a significant difference.
Alternative Solution: Using Newton’s second law of motion, F = ma, and inserting the
expression for the force that the electric field exerts on the electron, F = qE, we obtain
ma = qE
and therefore
a=
qE
m
The force on the electron is constant while it is between the capacitor plates, so its acceleration is also constant. Using a kinematics equation for the motion of the electron, we
obtain
v 2 − v02 = 2ad
v 2 − 02 = 2ad
√
v = 2ad
r
2qEd
v=
r m
2 (1.6 × 10−19 ) (2.1 × 106 ) (1.2 × 10−2 )
v=
9.11 × 10−31
v = 9.4 × 107 m/s
Note the similarity of the two solutions.
2. [4 marks] The figure shows an equilateral triangle, each side
of which has a length of 2.00 cm. Point charges are fixed to
each corner, as shown. The 4.00 µC charge experiences a net
force due to the charges qA and qB . This net force points vertically downward and has a magnitude of 405 N. Determine the
magnitudes and signs of the charges qA and qB .
Solution: Let Q represent the 4.00 µC charge, let FA represent the magnitude of the force
exerted by qA on Q, and let FB represent the magnitude of the force exerted by qB on Q.
Then
FA =
kqA Q
r2
and
FB =
kqB Q
r2
where r is the side length of the triangle.
Because the net force is vertically downwards, the horizontal component of the net force
on Q is zero, which means that
FB sin 30◦ − FA sin 30◦
FB
kqB Q
r2
qB
=0
= FA
kqA Q
=
r2
= qA
The magnitude of the net downward force on Q is
FB cos 30◦ + FA cos 30◦
FA cos 30◦ + FA cos 30◦
2FA cos 30◦
k|qA ||Q|
2·
cos 30◦
r2
= 405 N
= 405 N
= 405 N
= 405 N
405r2
|qA | =
2k|Q| cos 30◦
2
405 (2 × 10−2 )
2 (8.99 × 109 ) (4 × 10−6 ) cos 30◦
|qA | = 2.60 × 10−6 C
|qA | =
Thus, |qB | = |qA | = 2.60µC.
Because the force exerted by the other charges on Q is downward, qA and qB must both
be negative. Thus, qB = qA = −2.60µC.
3. [6 marks] In a large sample of hydrogen atoms, each hydrogen atom has its electron
excited into the n = 4 energy level. Determine all the possible wavelengths of the photons
emitted when the hydrogen atoms eventually return to their ground states.
Solution: There are six distinct transitions that electrons might make in making their way
from the n = 4 energy level to the n = 1 energy level; they are 4 −→ 3, 4 −→ 2, 4 −→ 1,
3 −→ 2, 3 −→ 1, and 2 −→ 1. One way to determine the corresponding wavelengths of
the emitted photons is to use the formula
!
1
1
− 2
hf = 13.6
n2f
ni
where the units for 13.6 are eV. Converting to SI units, we have
!
1
1
hf = 13.6 × 1.6 × 10−19
− 2
n2f
ni
Using f = c/λ, we can write the previous expression in terms of the wavelength as follows:
!
1
1
hc
= 13.6 × 1.6 × 10−19
− 2
2
λ
nf
ni
hc
λ=
13.6 × 1.6 ×
λ=
10−19
1
n2f
−
1
n2i
(6.63 × 10−34 ) (3.00 × 108 )
13.6 × 1.6 × 10−19 n12 − n12
f
i
−8
λ=
9.1406 × 10
1
− n12
n2
f
91.406
λ= 1
− n12
n2
f
(measured in m)
i
i
(measured in nm)
Inserting the appropriated values for the initial and final energy levels, we obtain the six
possible wavelengths:
λ4→1 =
λ4→2 =
λ4→3 =
λ3→1 =
λ3→2 =
λ2→1 =
91.406
1
− 412
12
91.406
1
− 412
22
91.406
1
− 412
32
91.406
1
− 312
12
91.406
1
− 312
22
91.406
1
− 212
12
= 97.5 nm
= 488 nm
= 1880 nm
= 103 nm
= 658 nm
= 122 nm
4. [4 marks] After 27 days, 96.1% of a sample of radioactive nuclei remains undecayed.
Determine the half-life of this species of nuclei.
Solution:
N = N0 2−t/T
N
= 2−t/T
N0
0.961 = 2−27/T
log (0.961) = log 2−27/T
27
log (0.961) = − · log (2)
T
27 log (2)
T =−
log (0.961)
T = 470 days
Note that the base of logarithms used in the calculations is irrelevant, provided that you
use the same base throughout.
5. [4 marks] In the Bohr model of the hydrogen atom, an electron moves around a proton
in a circular orbit of radius 5.3 × 10−11 m at a speed of 2.2 × 106 m/s. Consider the
moving electron as a small current loop and determine the magnetic field it produces at
the position of the proton.
Solution: First determine how much current corresponds to the electron’s motion. The
time T needed for the electron to orbit the nucleus once is related to the radius r of the
orbit and the electron’s speed v by
T =
2πr
v
The definition of current is the amount of charge flowing past a point per unit time; in
the case of the electron, it flows past a point on the circumference of its orbit once per
period. Thus, the current corresponding to the electron’s flow is
e
T
ev
I=
2πr
I=
The magnetic field at the centre of a current loop is
B=
µ0 I
2r
and so the magnetic field produced by the electron’s flow at the centre of its orbit is
µ0 ev
·
2r 2πr
µ0 ev
B=
4πr2
(4π × 10−7 ) (1.6 × 10−19 ) (2.2 × 106 )
B=
4π (5.3 × 10−11 )2
B = 12.5 T
B=
6. [4 marks] A distant galaxy is simultaneously rotating and
receding from the earth. As the drawing shows, the galactic centre is receding from the earth at a relative speed of
uG = 1.6×106 m/s. Relative to the centre, the tangential speed
is 0.4 × 106 m/s for locations A and B, which are equidistant
from the centre. When the frequencies of the light coming from
regions A and B are measured on earth, they are not the same
and each is different from the emitted frequency of 6.200 × 1014
Hz. Find the measured frequency for the light from
(a) region A.
(b) region B.
Solution: The speed of region A relative to the Earth is
1.6 × 106 + −0.4 × 106 = 1.2 × 106 m/s
The speed of region B relative to the Earth is
1.6 × 106 + 0.4 × 106 = 2.0 × 106 m/s
Each region recedes from the Earth, so to determine the observed light frequency, we use
the lower sign in the Doppler shift formula; thus,
vrel fo = fs 1 −
c
The light from region A is observed with a frequency
1.2 × 106
14
1−
foA = 6.200 × 10
3.00 × 108
foA = 6.175 × 1014 Hz
The light from region B is observed with a frequency
2.0 × 106
14
1−
foB = 6.200 × 10
3.00 × 108
14
foB = 6.159 × 10 Hz
7. [4 marks] Determine the current through each resistor.
10 Ω
10 V
30 Ω
20 Ω
40 V
10 Ω
30 Ω
Solution: I’ll choose to label the unknown currents in the circuit as follows:
10 Ω
30 Ω
I1
I3
10 V
20 Ω
I2
I3
10 Ω
30 Ω
I1
I3
40 V
I’ll solve this problem using Kirchhoff’s laws. Using Kirchhoff’s loop law for the left loop,
10 − 10I1 − 20I2 − 10I1 = 0
which simplifies to
20I1 + 20I2 = 10
2I1 + 2I2 = 1
(1)
Using Kirchhoff’s loop law for the large loop,
10 − 10I1 − 30I3 + 40 − 30I3 − 10I1 = 0
which simplifies to
20I1 + 60I3 = 50
2I1 + 6I3 = 5
(2)
Using Kirchhoff’s junction law for either of the two junctions, we obtain
I1 = I2 + I3
(3)
We have three independent equations for the three unknown currents, so the problem is
solvable. One way to solve the system of equations is to substitute the expression for I1
from equation (3) into equations (1) and (2) to eliminate I1 . Doing this with equation (1)
and simplifying, one obtains
2(I2 + I3 ) + 2I2 = 1
4I2 + 2I3 = 1
(4)
Doing this with equation (2) and simplifying, one obtains
2(I2 + I3 ) + 6I3 = 5
2I2 + 8I3 = 5
−4I2 − 16I3 = −10
(5)
Adding equations (4) and (5), one obtains
−14I3 = −9
9
A
I3 =
14
Substituting the value of I3 into equation (4) and solving for I2 , one obtains
9
4I2 + 2
=1
14
9
4I2 + = 1
7
9
4I2 = 1 −
7
7 9
4I2 = −
7 7
2
4I2 = −
7
2
I2 = −
(4)(7)
1
I2 = − A
14
The fact that the calculated value of I2 is negative indicates that our supposed direction
for I2 in the diagram is incorrect; current actually flows upwards in the 20-Ω resistor, not
downwards. Finally, substituting the values of I2 and I3 into equation (3) gives us the
value of I1 :
I1 = I2 + I3
9
1
I1 = − +
14 14
8
I1 =
14
4
I1 = A
7
8. [4 marks] A conducting rod slides down between two frictionless vertical copper tracks
at a constant speed of 4.0 m/s perpendicular to a 0.50-T magnetic field. The resistance of
the rod and tracks is negligible. The rod maintains electrical contact with the tracks at
all times and has a length of 1.3 m. A 0.75-Ω resistor is attached between the tops of the
tracks. Determine the mass of the rod.
Solution: Because the rod is moving in a magnetic field, an emf is induced in the rod:
E = B`v
E = (0.5) (1.3) (4)
E = 2.6 V
Thus, the current in the rod is
E
R
2.6
I=
0.75
I = 3.47 A
I=
Because the rod carries a current, the magnetic field exerts a force on it. Because the rod
falls at a constant speed, the force on the rod due to the magnetic field balances the force
on the rod due to gravity. Thus,
mg = I`B
I`B
m=
g
2.6
(1.3) (0.5)
m=
·
0.75
9.8
m = 0.23 kg
9. [6 marks] In lectures we discussed four main sources of evidence supporting the photon hypothesis. Choose three of these sources of evidence and briefly explain how each
supports the photon hypothesis.
Solution: The four main sources of evidence for the photon hypothesis that we discussed
in lectures are, (a) Planck’s introduction of energy quantization in his analysis of blackbody
radiation, (b) Einstein’s explanation of the photoelectric effect, (c) the Compton effect, and
(d) the formation of double-slit interference patterns using low-intensity particle beams.
(a) Planck was able to derive a family of formulas that fit experimental blackbody data
by assuming that the energy in “fundamental vibrators” in a glowing object is collected
up in bundles of certain definite sizes before being radiated away by electromagnetic radiation. He hypothesized that each bundle of energy is proportional to the frequency of
the vibration; that is, E = hf . This is consistent with the photon hypothesis, although
Planck himself did not go this far.
(b) Einstein did take the bold step of hypothesizing that electromagnetic radiation exists
as particles, which were later called photons. He brilliantly used the photon hypothesis
to explain the photoelectric effect. Because it was not possible for others to explain the
photoelectric effect using the wave theory of electromagnetic radiation, this was a powerful
argument in favour of the photon hypothesis.
(c) If light really exists as particles, then these particles ought to carry momentum as
well as energy, and one ought to be able to analyze the scattering of light from particles
in the same way that collisions between billiard balls are analyzed. Compton did such
an analysis, treating electromagnetic waves as photons that carry momentum and energy.
His results were confirmed by his own experiments, which therefore provided convincing
support for the photon hypothesis.
(d) If light is made of particles, why then does light form interference patterns as if it were
really waves? Well, if you run light through a double-slit experiment but decrease the
intensity of the light to a very low level, you can see how the interference pattern develops
bit-by-bit. At first you see just a few scattered dots on the screen, which is consistent
with light being localized as particles. As time passes, more and more dots appear on the
screen, consistent with light being made up of photons. Only after a sufficient amount of
time passes does the collection of dots coalesce into what looks like a wave interference
pattern.
This highlights another puzzle (wave-particle duality), but the step-by-step development
of the pattern on the screen is clear support for the photon hypothesis.
10. (a) [2 marks] What is a nuclear chain reaction? Describe how such a chain reaction
might lead to an explosion, or how it might lead to a controllable, steady release of
energy. Which factors might decide which occurs?
(b) [2 marks] Explain schematically how a nuclear reactor works to generate electricity.
What is a neutron moderator, and why is one needed in a nuclear reactor? What is
a neutron absorber, and why is one needed in a nuclear reactor?
Solution: (a) Using uranium as an example, when a neutron strikes a uranium nucleus,
there is a possibility that the uranium nucleus may break up into two large fragments
and two or three neutrons (this is called nuclear fission). The neutrons emerge at high
speeds, and they may strike other uranium nuclei, stimulating more fission reactions. If
this continues, the number of free neutrons flying around (and, therefore, the number of
subsequent fissions stimulated) may increase with every step of the process. This is called
a chain reaction.
If it occurs that most of the two or three neutrons released per fission goes on to stimulate another fission, this results in an uncontrolled release of energy, amounting to an
explosive reaction. If it occurs that at most one of the neutrons released per fission goes
on to stimulate another fission, this results in a release of energy that may be stable and
controllable.
Which alternative occurs depends on how closely packed the fissionable nuclei are, and
also on whether there are substances around that might absorb some of the neutrons or
slow some of them down. Neutron absorbers may be able to absorb a sufficient number of
neutrons to prevent explosive reactions and help a nuclear chain reaction to be controllable.
Substances that slow neutrons as they pass through are called neutron moderators; slow
neutrons are much more likely to stimulate fission than fast neutrons.
(b) Nuclear reactors release energy that is used to boil water. The resulting steam is
directed at high pressure to turn a turbine that sits in a magnetic field. Coils of wire are
attached to the turbine, and as they turn within the magnetic field, electricity is generated
according to Faraday’s law of induction.
In the core of the reactor are the fuel rods, which contain fissionable material such as
uranium. When a uranium nucleus spontaneously decays, it releases a neutron that may
strike another nucleus and stimulate fission, releasing two or three more nuclei. The
neutrons pass through a moderator that is placed between the fuel rods, allowing them
to slow down and then hit other fuel rods. Because the neutrons have slowed down, the
probability that they will stimulate fission when they are incident on another nucleus is
much increased.
Between the fuel rods are also control rods, which contain neutron absorbers. When the
control rods are positioned ideally, the number of neutrons absorbed leaves about one
neutron per fission available to hit other uranium nuclei and stimulate fission, which can
lead to a stable, controlled release of energy. (The fission fragments are released with a
significant amount of kinetic energy, which is transferred as heat to the coolant water that
surrounds the reaction vessel.) The coolant water then passes through a heat-exchanger,
where its heat is transferred to water that will drive the electrical generator.
The nuclear reactor is carefully monitored, with operators continually adjusting the positioning of the control rods and other factors to keep the reactor running at the desired
output of electrical energy.
11. (a) [1 marks] Briefly explain what the equivalence principle is. (In lecture we discussed
two aspects of the equivalence principle; discuss whichever one you prefer.)
(b) [2 marks] Briefly explain the basic ideas of Einstein’s theory of gravity. That is, in
Einstein’s perspective, what are the sources of gravitational fields, how do gravitational fields manifest, and how do particles respond to gravitational fields?
(c) [2 marks] Briefly discuss one of the successful predictions of Einstein’s theory of
gravity, and briefly outline the experimental support for it.
Solution: (a) There are two, more-or-less equivalent versions of the equivalence principle.
The first version states that inertial mass and gravitational mass are equivalent. That
is, the property of a body that responds to and is the source of gravitational force is
numerically identical to the property of a body that resists acceleration according to
Newton’s second law of motion. The second version of the equivalence principle states
that a gravitational field is locally equivalent to a reference frame that has an appropriate
value of acceleration.
(b) The slogan is, “Matter (and energy, etc.) tells spacetime how to curve, and spacetime
tells particles how to move.”
In a little more detail, the density of mass, energy, and related factors (pressure, etc.) is
the source of the gravitational field. Einstein’s equations describe mathematically how the
curvature of spacetime is related to the density of mass, energy, etc., in the universe.
Freely moving particles (including photons, particles of light) follow the contours of spacetime as they move. That is, free particles move along geodesics, which are the straightest
possible curves on the surface of spacetime.
(c) In lectures we discussed a number of experimental predictions of Einstein’s theory
of gravity, all of which have been repeatedly and accurately confirmed. You might have
chosen to focus your response on one of the three classical tests of general relativity, these
being the perihilion shift of the orbit of Mercury, the bending of starlight grazing the
surface of the Sun, and gravitational redshift. Alternatively, you might have chosen to
discuss gravitational time dilation (which has critical implications for the operation of the
Global Positioning System (GPS), or perhaps the event horizons of black holes.
Detailed explanations of the experiments and observations confirming all of these phenomena can be found in the lecture notes.