Physics 6C
Final Practice Solutions
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Use the following information for problems 1 and 2.
A beam of white light with frequency between 4.00 x 1014 Hz and 7.90 x 1014 Hz is incident on a
sodium surface, which has a work function of 2.28 eV.
1) What is the range of frequencies in this beam of light for which electrons are ejected from the
sodium surface?
An easy way to do this one is to find the frequency of a photon that has an energy of 2.28 eV.
Ephoton  h  f
2.28eV  (4.14  1015 eV  s)  (f )
f  5.5  1014 Hz
Now we just have to realize that any photon with a higher frequency will have more energy,
and will also eject electrons.
So the answer is (d).
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Use the following information for problems 1 and 2.
A beam of white light with frequency between 4.00 x 1014 Hz and 7.90 x 1014 Hz is incident on a
sodium surface, which has a work function of 2.28 eV.
2) What is the maximum kinetic energy of the photoelectrons that are ejected from this surface?
We use the most energetic photon available – the one with the highest frequency. Its
energy will be higher than the work function, and the difference will be the leftover KE.
KEmax  h  f  
KEmax  (4.14  1015 eV  s)  (7.9  1014 Hz)  2.28eV
KEmax  0.99eV
Convert to Joules:
KEmax  (0.99eV)  (1.6  1019
J
eV
)  1.6  1019 J
Answer (b)
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3) When light of a particular wavelength and intensity shines on the surface of a given metal, no
electrons are ejected. Which of the following changes will most likely cause the emission of
photoelectrons?
a) Decreasing the frequency of the light
b) Increasing the intensity of the light
c) Decreasing the wavelength of the light
d) Increasing the wavelength of the light
In order for electrons to be ejected, the photons have to have more energy. To
do this we need to either increase the frequency or decrease the wavelength.
Changing the intensity will not change the energy of individual photons.
Answer (c).
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4) What is the De Broglie wavelength for a Helium atom (atomic mass=4.002603 u) traveling at a
speed of 103 m/s? Use the conversion factor 1 u = 1.66054 x 10-27 kg.
This one is just a formula:
h
6.63  1034 J  s


m  v (4.002603u)  (1.66054  1027
kg
u
)  (103 ms )
  1.0  1010 m
Answer (b). You may have to round up carefully on this one.
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5) The neutral pion (π0) is an unstable particle produced in high-energy particle collisions. Its mass
is 264 times that of the electron, and it exists for an average lifetime of 8.4x10-17s before
decaying into 2 gamma-ray photons. Find the uncertainty in the mass of the pion, expressed as
a fraction of its mass.
Δm/m is approximately:
We use the uncertainty principle for this one:
E  t 
h
2
6.63  1034 J  s
E  (8.4  10 s) 
2
E  1.256  1018 J
17
At this point use E=mc2:
1.256  1018 J  m  c 2
m  1.4  1035 kg
Divide by the mass of the pion to get the ratio we need:
m
1.4  1035 kg

 5.8  10 8
 31
mpion 264  9.1  10 kg
Answer (c).
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6) You and a friend travel through space in identical spaceships. Your friend informs you that he
has made some length measurements and that his ship is 150m long and yours is 120m long.
What is the relative speed of the two ships?
Use the length contraction formula:
L  L 0 1 
v2
c2
120m  (150m) 1 
0.8  1 
v2
c2
0.64  1 
v2
c2
v2
c2
v2
c2
 0.36
v  0.6c
Answer (c).
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7) A laser is used to move the electron in a hydrogen atom from the n=2 to the n=4 state. What is
the wavelength of the light used to excite this electron?
The energy of the absorbed photon is the difference in the energies of the excited states.
 13.6eV
n2
 13.6eV  13.6eV
E4  E2 

 2.55eV
42
22
En 
Now find the wavelength of a photon with this energy:
hc

(4.14  1015 ev  s)  (3  108 ms )
2.55eV 

7
  4.87  10 m  490nm
Ephoton 
Answer (b).
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8) Assume the excited hydrogen atoms in problem 7 will emit photons until they fall back to the
ground state. What is the minimum wavelength of the emitted photons?
Since we want the minimum wavelength, we want the photon with the maximum energy.
That will be the one that drops all the way to the ground state (i.e. from n=4 to n=1).
E1  E4   13.6eV 
 13.6eV
 12.75eV
42
Now find the wavelength of a photon with this energy:
hc

(4.14  1015 ev  s)  (3  108 ms )
12.75eV 

8
  9.7  10 m  97nm
Ephoton 
Answer (b).
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9) An object is 4m away from a wall. You are to use a concave mirror to project an image of the
object on the wall, with the image 2.25 times the size of the object. What should the radius of
curvature of the mirror be?
a) 1.1m
b) 2.2m
do
4m
c) 4.4m
di
d) 5.5m
Since the image is real, it will be inverted, so we should have a magnification m=-2.25.
We can use the magnification formula to find the ratio of the object and image, then use the
picture to find a value for do. Plug in to find focal length. Radius is twice the focal length.
−𝑑𝑖
= −2.25
𝑑𝑜
𝑑𝑖 = 2.25𝑑𝑜
𝑚=
𝑑𝑖 = 4𝑚 + 𝑑𝑜 = 2.25𝑑𝑜
4𝑚 = 1.25𝑑𝑜 → 𝑑𝑜 = 3.2𝑚
1
1
1
1
1
1
1
1
=
+ =
+
=
+
→ = +0.45
𝑓 𝑑𝑜 𝑑𝑖 𝑑𝑜 2.25𝑑𝑜 3.2𝑚 2.25(3.2𝑚) 𝑓
𝑓 = +2.2𝑚
𝑅 = +4.4𝑚
answer c)
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10) A 239Pu (plutonium) nucleus typically decays by emitting an α-particle with an energy of about
5.1 MeV. What is the resulting daughter nucleus?
Alpha-decay ejects a helium nucleus from the nucleus of the large atom, leaving
it with 2 fewer protons, and 4 fewer nucleons overall. So we need to use the
periodic table and find the element that has 2 fewer protons than plutonium.
This is uranium. The number of nucleons should be 4 fewer, so we get
235U.
Answer (a).
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11) Nucleus X has 10 times as many nucleons as nucleus Y. Compared to Y, X has
a) a larger radius and about the same density
b) a larger radius and a smaller density
c) a smaller radius and about the same density
d) a larger radius and a larger density
We have a formula for nuclear radius:
1
R  (1.2  1015 m)  (A 3 )
So nucleus X should have a larger radius since it has a larger value of A.
However, nuclear density is about the same for all nucleons (we did this in class).
Answer (a).
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12) Originally the decay rate of a sample was 16 decays per second. Now it has a rate of 2 decays
per second. The half-life is 5 years. How many years have passed?
If you notice that the numbers come out perfectly, then you can do this one
without much calculating. The decay rate has dropped from 16 to 2. This is a
factor of 8. Since 23=8, we know the time should be 3 half-lives, or 15 years.
Answer (d).
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13) An ancient bone has been recovered from an archeological dig. It is determined that the ratio
of the isotope carbon-14 to total carbon is only 20% of the ratio currently present in the
atmosphere. The half-life of carbon-14 is 5730 years. How old is the bone?
We will need to use our exponential decay formula for this one.
The decay constant comes from the half-life:
N(t)  N0  et

ln(2)
 1.21  10 4
T12
1
yr
(20%)  N0  N0e t
0.2  e t
ln(0.2)  (1.21  10 4
1
yr
) t
t  13,300yr
Answer (c).
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14)
is an unstable isotope of nitrogen with a half-life of about 7 seconds. When it undergoes
decay an electron, an antineutrino, and one other particle are emitted. What is the other
particle?
16
-
7N
The other particle will be the daughter nucleus, which should have one more
proton than the original nitrogen atom. From looking at the periodic table, we
find that it must be oxygen.
Answer (c).
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15) A light ray in water is incident at an angle of 22.8° on an oil film with thickness 560 nm and
index of refraction 1.45. Does the transmitted light ray bend toward or away from the normal?
Assume that the index of refraction of water is 1.33.
a) Toward
b) Away
c) The ray does not bend at all.
d) The ray is completely reflected.
The ray is entering a higher index material, so it should bend toward the normal.
Snell’s Law will calculate the angle if we need it.
𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2
1.33 sin 22.8° = 1.45 sin 𝜃2
Oil n2=1.45
𝜃2 = 20.8°
Water n1=1.33
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16) An oil film floats on water. The refractive index of the oil is 1.4 and the index for water is
1.33. The film appears blue because it reflects light with wavelength 400 nm very well.
Calculate the minimum thickness of the oil film.
1
Ray 1 reflects from a higher index, but ray 2 reflects
from a lower index, so there is a relative phase shift.
Our formulas are:

m  0  2  t  destructive
n

(m  1 )  0  2  t  constructive
2
n
2
Air n=1
Oil n=1.40
Water n=1.33
We use the constructive formula, with
m=0 to get the thinnest possible film:
400nm
 2t
1.4
 71nm
(0  12 ) 
tmin
Answer (a).
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17) A single-slit experiment is performed with light of wavelength of 500 nm, and the first dark
fringe appears at a distance of 4cm from the central bright maximum. If this experiment is
now performed with the same setup, but underwater, how far from center will the first dark
fringe be? Assume the index of refraction of water is 1.33.
This is a single-slit experiment, so we can use the formula for the position of the
dark fringes. Use m=1 to get the first dark fringe from center.
d sin( )  m
This will get us the angle. Now use a right triangle to get distance on the screen.
y  R tan( )
We can combine these formulas if we realize that for small angles, tan(θ)~sin(θ).
y 
mR
d
Before we plug in the numbers, realize that the only difference between the
experiments will be the wavelength of the light. Underwater, the wavelength will
be shortened. So we can set up a ratio to solve for the distance.
ywater

 water
y air
air
ywater 
y air
4cm

 3cm
n
1.33
Answer (b).
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18) A double-concave lens is made from glass with index of refraction n=1.6. The radius of
curvature is 90cm for each side. Find the focal length of this lens when used in air.
a)
50cm
b)
-90cm
c)
-75cm
d)
-50cm
Assuming the light comes from the
left, as shown in the diagram, we
will get a negative value for R1 and
a positive for R2.
The convention is that the incoming
light side is negative, and the
outgoing side is positive.
Just plug into the formula:
1
= 𝑛−1
𝑓
1
1
−
= 1.6 − 1
𝑅1 𝑅2
1
1
−1
−
= (0.6)
−90𝑐𝑚 90𝑐𝑚
45𝑐𝑚
f = −75cm
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19) The sun produces energy by nuclear fusion reactions, in which matter is converted to energy.
The rate of energy production is 3.8 x 1026 Watts. How many kilograms of mass does the sun
convert to energy each second?
Recall that a Watt is a Joule/Second.
Convert energy to mass using Einstein’s formula:
E  m  c2
3.8  1026 J  (m)  (3  108 ms )2
m  4.2  109kg
Answer (a).
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20) Calculate the mass defect for the decay of 84Be into two alpha particles.
The mass of 84Be is 8.005305 u, and the mass of an alpha particle is 4.00148 u.
Use the conversion factor 1 u = 931.5 MeV
We need to find the difference between the mass of the original Be nucleus (don’t forget to
account for the electrons!), and the mass of the 2 alpha particles that result from the decay:
Beryllium nucleus




m  8.005305u  4(0.000549u)  2(4.00148u)  0.000149u
m  (0.000149u)  (931.5 MeV
u )  0.14MeV
Answer (c).
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21) Polarizer 1 has a vertical transmission axis. Polarizer 2 has a transmission axis tilted 45°
from the vertical, and polarizer 3 has a horizontal transmission axis. The light incident on
polarizer 1 is polarized vertically. If polarizer 2 is removed, what will be the effect on the
intensity of light at point P?
a) the intensity will be zero (no light will emerge from polarizer 3)
b) the intensity will increase by a factor of 2
c) the intensity will decrease, but will be greater than zero
d) the intensity will be unchanged
When polarizer #2 is removed, it leaves #1 and #3 next to each other. Their transmission
axes are perpendicular to each other, so no light gets through.
Answer (a).
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22) At a point in space an electromagnetic wave is propagating in the negative y direction. At
a certain instant, the magnetic field of the wave at this point is in the positive x direction.
At this point and this instant, the electric field of the wave is
a) in the positive z direction
b) in the positive x direction
c) in the negative y direction
d) in the negative z direction
The right-hand-rule for a plane EM wave is the following: E, B and v are mutually
perpendicular. Start with your fingers along E, bend them toward B, and your thumb is
along the velocity (propagation direction).
In our problem we know that the E-field
must be in the z-direction (since B and v
are already in x or y). Try it with E-field
in the +z direction – you get velocity in
the +y direction. It must be the other
way since velocity comes out in the –y
direction.
So the answer is E-field in –z direction
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23) An astronaut traveling with a speed v relative to Earth takes her pulse and measures 70
beats per minute. Mission control on Earth, which monitors her heart, observes a rate of 43
beats per minute. What is the astronaut’s speed relative to Earth?
a)
0.5c
b)
0.7c
c)
0.8c
d)
0.9c
We will use the time dilation formula for this one. It may help to convert the
given measurements into time intervals first, although this is not necessary.
1 𝑚𝑖𝑛
70 𝑏𝑒𝑎𝑡𝑠
∆𝑡 =
60 𝑠𝑒𝑐
𝑠𝑒𝑐
= 0.86
1 𝑚𝑖𝑛
𝑏𝑒𝑎𝑡
1 𝑚𝑖𝑛
43 𝑏𝑒𝑎𝑡𝑠
60 𝑠𝑒𝑐
𝑠𝑒𝑐
= 1.40
1 𝑚𝑖𝑛
𝑏𝑒𝑎𝑡
∆𝑡0
𝑣2
1− 2
𝑐
1.40𝑠𝑒𝑐 =
0.86𝑠𝑒𝑐
1−
𝑣2
→
𝑣 2 0.86𝑠𝑒𝑐
𝑣2
1− 2 =
→ 1 − 2 = 0.61 → 𝑣 = 0.8𝑐
𝑐
1.40𝑠𝑒𝑐
𝑐
𝑐2
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