PHY1160C
Final Exam
July 15, 1997
Fel = k
Qq
r2
E =F
q
k = 9 x 109 N m 2/C 2
e = 1.6 x 10–19 C
Q
E=k
r2
∆PEel = k Qq (r1 – r1 )
f
i
AB
∆VAB = ∆PE
q
C=
Q
V
Q2
PEcap = 1 QV = 1 CV2 = 1
2
2
2 C
1 = 1 + 1 + 1
Ceq C1 C2 C3
Ceq = C1 + C 2 + C 3
R=V
I
P = IV = I2R = V 2/R
1 = 1 + 1 + 1
Req R1 R2 R3
Req = R1 + R 2 + R 3 + . . .
Fmag = |q| v B sin ϕ
F = I B l sin ϕ
µ I
B= o
2πr
τ = N I A B sin θ
E=–N∆ Φ
∆t
f = ω/2π
E = – L ∆i
∆t
v = Vo sin ωt
Z = R2 + (XL - XC)2
F =l I x B
µ I
Bcenter = o
2a
B = µo n I
E=–∆ Φ
∆t
Φ = B A cos θ
Φ = BA cos θ = BA cos ωt
Eo = N ω B A
E=vBl
E = Eo sin ωt
i = I f [ 1 – e–(R/L)t ]
XC = 1
2π f C
V = IZ
p = mv
F = m v2/ r
KE = (1/ 2) m v2
For reflection,
θ inc = θ ref
F = ma
Fmag = q v x B
V1 = V2
N1 N2
X L = 2π f L
1
fo = 1
2π
LC
F=qE
PE = mgy
Final - p 1 -
For refraction,
image equation
apparent depth
Young’s Double Slit:
Diffraction Grating:
Diffraction by a slit:
Optical Resolution:
Thin Films:
n 1 sin θ 1 = n2 sin θ 2
n = c/v
h
d
M= i =– i
ho
do
1 =1 +1
f di do
di = do (n1/n 2)
∆ = d sin θ
d sin θ = m λ,
maxima for m = 0, 1, 2, . . .
d sin θ = (m + 1/ 2) λ, minima for m = 0, 1, 2, . . .
∆ = d sin θ
d sin θ = m λ,
maxima for m = 0, 1, 2, . . .
∆ = D sin θ
First minimum for ∆ = D sin θ = λ
sin θ = λ/D
θ ≈ λ/D
Minima for ∆ = D sin θ = m λ, m = 1, 2, 3, . . .
θ ≈ 1.22( λ/D )
λ’ = λ/n
2t = m λ’ = m(λ/n); max, m = 0, 1, 2, . . . n 1 < n2 < n3
2t = (m+1/ 2) λ’ = (m+ 1/ 2)(λ/n);
min, m = 0, 1, 2, . . . n 1 < n2 < n3
Simple Magnifier:
2t = m λ’ = m(λ/n); min, m = 0, 1, 2, . . . n 1 < n2 > n3
2t = (m+1/ 2) λ’ = (m+ 1/ 2)(λ/n);
max, m = 0, 1, 2, . . . n 1 < n2 > n3
M ang = r / f = 25 cm / f
Mtot = –
Microscope:
L – feye 25 cm
fobj
feye
Telescope:
f
Mang = θ = – obj
feye
θ’
c = 3.00 x 108 m/s
xA = xB + v t B
1 – (v 2/c2)
h = 6.626 x 10–34 Js = 4.14 x 10–15 eV s
t + (v/c2) xB
tA = B
1 – (v 2/c2)
L = Lo 1 –
(v 2/c2)
∆t =
∆to
1 – (v 2/c2)
Final - p 2 -
m=
mo
1 – v2 /c2
v xMA =
v xMB + v
1 + (vv xMB /c2)
E=hf
E = m c2
KEmax = h f – Wo
l = h/p = h / mv
(Dx) (Dpx ) ≈ h
for SHO, Eo = h f / 2
1 =R 1 – 1
n = 3, 4, . . .
n2
λ
22
R = 1.097 x 107 m–1
2
Etot = – 1 k Ze
r
2
En = – Z2 13.6 eV
n2
Lz = ml h
2π
rn =
L = l ( l + 1) h
2π
S z = ms h
2π
P
A
Z
P
A
Z
P
—>
—>
—>
A–4
Z–2
D
A
Z+1
D
A
Z–1
D
N = No e– λ t
4
He
+ 2
n2 h2
4π 2mekZe2
n = 1, 2, 3, . . .
Sz = ± 1 h
2 2π
Subshells: l = 0, 1, 2, 3, . . .; s, p, d, f, g, h, . . .
u
electron
0.00054858
proton
1.007276
neutron
1.008665
hydrogen
1.007 825
A
Z
DE = Ei – Ef = h f
1 u = 931 MeV/c2
Mev/c2
kg
0.511
9.1094 x 10–31
938.27
1.67262 x 10–27
939.57
1.67493 x 10–27
Q = ( mP – m D – m a) c2
+ b– + ν
Q = ( mP – m D ) c2
+ b+ + n
Q = ( mP – m D – 2 m b ) c2
T1/2 = 0.693/ λ
fλ=v
For every question, also consider, as a possible answer,
E) none of the above.
Final - p 3 -
1. The force between two charges of Q1 = 1 µC and q2 = – 2 µC when they are separated by a
distance of r = 3 m is
A) 1 x 10–2 N
B) 2 x 10–3 N
C) 3 x 10–3 N
D) 3 x 10–4 N
Qq (9 x 10 9) (1 x 10 –6) (2 x 10 –6)
F=k
=
N
r2
(3)2
F = 2 x 10–3 N
2. The electric field at a distance of 3 m from an electric charge of Q = 2 µC is
A) 1 x 102 N/ C
B) 2 x 103 N/ C
C) 3 x 103 N/ C
D) 6 x 103 N/ C
E=k
Q (9 x 10 9) (2 x 10 –6) N
=
C
r2
(3)2
E = 2 x 103 N/ C
3. The electric potential at a distance of 3 m from an electric charge of Q = 2 µC is
A) 1 x 102 V
B) 2 x 103 V
C) 3 x 103 V
D) 6 x 103 V
Q (9 x 10 9) (2 x 10 –6)
V=kr =
V
(3 )
V = 6 x 103 V
4. What is the potential difference between two points if 6 µJ of work is required to move 12 µC
of charge from one point to the other?
A) 0.5 V
B) 2.0 V
C) 6.0 V
D) 12.0 V
∆PEel = W = 6 µJ
∆V = ∆PEel = 6 µJ = 0.5 J = 0.5 V
q
12 µC
C
∆V = 0.5 V
Final - p 4 -
5. Calculate the equivalent resistor for this circuit if the resistors have the following values:
R1 = 500 Ω, R 2 = 750 Ω, and R3 = 1500 Ω.
A) 158 Ω
B) 250 Ω
C) 2 750 Ω
D) 3 250 Ω
V
R1
R2
R
3
These resistors are connected in parallel so the equivalent resistance, Req, is given by
1 = 1 + 1 + 1
Req R1 R2 R3
1 /Req = 1 / 500 Ω + 1 / 750 Ω + 1 /1500 Ω
1 /Req = (0.002 + .00133 + .00067) / Ω
1 /Req = 0.004 / Ω
Req = 250 Ω
6. Calculate the equivalent resistor for this circuit if the resistors have the following values:
R1 = 500 Ω, R 2 = 750 Ω, and R3 = 1000 Ω.
A) 230 Ω
B) 500 Ω
C) 1 250 Ω
D) 2 250 Ω
R1
R3
R2
V
These resistors are in series so the equivalent resistor is
Req = R1 + R2 + R3
Req = 500 Ω + 750 Ω + 1000 Ω
Req = 2 250 Ω
Final - p 5 -
7. A rectangular loop has dimensions of 2.0 cm by 3.0 cm and is initially between the poles of a
large magnet, where the magnetic field strength is B = 0.050 T. The loop is oriented so the
normal to the plane of the coil lies parallel to the direction of the magnetic field. The coil is
removed from that region to a region with no magnetic field in 0.15 seconds. What is the
average induced emf?
A) E = 0.000 1 V = 1 mV
B) E = 0.000 2 V = 2 mV
C) E = 0.000 3 V = 3 mV
D) E = 0.000 5 V = 5 mV
Φo = B A = (0.05 T)(0.02 m)(0.03 m)
Φo = 0.000 030 T m2 = 0.000 030 W = 3 x 10–5 W
ΦFf = 0
∆Φ f = Φf – ΦFo = – 3 x 10–5 W
E=–
E=
∆ Φ
∆t
3 x 10–5 W
0.15 s
E = 0.000 2 V = 2 mV
8. Ordinary household electricity is said to be 120 volts. This means the rms voltage is 120
volts. Therefore, the peak voltage is
A) 85 V
B) 120 V
C) 170 V
D) 240 V
Vrms = 1 Vo ; V o = 2 Vrms = (1.414)(120V)
2
V o = 170 V
9. An electric light bulb, connected to an ordinary 120 V AC wall outlet, uses 60 W of power.
What is the resistance of the light bulb?
A) R = 120 Ω
B) R = 240 Ω
C) R = 360 Ω
D) R = 480 Ω
P = IV
60 W = (I) (120 V)
I = 60 W = 0.5 A
120 V
I rms = 0.5 A
Final - p 6 -
I =V
R = V = 120 V = 240 Ω
R , or
I
0.5 A
R = 240 Ω
10. When 634-nm light passes through a single slit 0.1 mm wide and forms a diffraction
pattern on a screen 1.25 m away, what is the distance from the central maximum to the
center of the first minimum?
y=?
λ = 634 nm
θ
1.25 m
D = 0.1 mm
For a single-slit diffraction pattern, the first minimum occurs for
D sin θ = λ
or
sin θ = λ/D
sin θ = 634 nm/0.1mm
sin θ = (634 x 10–9 m)/(0.1x 10–3 m)
sin θ = 6.34 x 10–3 = 0.00634
The distance y is given by
tan θ = y/L
where L is the distance from the slit to the screen, L = 1.25 m
y = L tan θ
For such small angles
tan q ≈ sin θ
y = L sin q
y = (1.25 m)(0.00634)
y = 7.93 mm
A) 2 mm
B) 4 mm
C) 8 mm
D) 16 mm
Final - p 7 -
11. An astronaut in a space lab focuses a 60 mm focal length camera lens to form a sharp
image of Earth. What is the distance between the film and the lens?
Earth is essentially infinitely far away from the lens so do = ∞. We can then calculate the
initial image distance di1.
1= 1 +1
f do di
1
=1 + 1
60 mm ∞ di1
di1 = 60 mm
A) 15 mm
B) 30 mm
C) 60 mm
D) 120 mm
12. The astronaut of the previous problem now moves the 60 mm focal length lens in order to
focus on a fellow astronaut, 1.0 m away? What is the distance between the film and the lens
now?
We calculate the image distance di2 when the object distance is do = 1.0 m = 1,000 mm,
1= 1 +1
f do di
1
1
=
+ 1
60 mm 1,000 mm di2
1 =
1
1
–
di2 60 mm 1,000 mm
1 = 0.01667 – 0.001
= 0.01567
mm
mm
di2
di2 = 63.82 mm
A) 3.8 mm
B) 56.2 mm
C) 60.0 mm
D) 63.8 mm
Final - p 8 -
13. A SCUBA diver, underwater, shines a light up toward the smooth surface of the water
with an angle of incidence of 35°. At what angle does the light leave the water? nwater = 1.33,
n air = 1.0
A) 20°
B) 35°
C) 50°
D) 65°
air
θ2
n2
n1
θ1
water
We begin with Snell’s law,
n1 sin θ 1 = n2 sin θ 2
(1.33)(sin 35°) = (1.00) sin θ 2
(1.33)(0.5736) = (1.00) sin θ 2
sin q2 = 0.7269
θ 2 = 49.7° ≈ 50°
14. If you look at your toes while standing in 1.0 m of water, how far below the surface will
they appear to be? The index of refraction for water is 1.33.
A) 65 cm
B) 75 cm
C) 100 cm
D) 133 cm
n = 1.00
n = 1.33
1m
The apparent depth is given by
1
di = do n
n
2
di = (1.0 m) 1.00
1.33
di = 0.75 m = 75 cm
Final - p 9 -
15. A patient's eye can focus only on objects beyond 100 cm. What is the focal length of the
contact lens needed to correct this problem?
A) f = – 20 cm
B) f = – 33 cm
C) f = + 33 cm
D) f = + 50 cm
This patient is farsighted or hyperopic. This patient has trouble reading a book at a
comfortable distance—such as 25 cm—so we must find a lens that will take an object at 25 cm
(do = 25 cm) and create an image (a virtual image) at 100 cm (di = – 100 cm).
virtual image at
real object at
d = – 100 cm
do = 25 cm
i
Near point:
25 cm is used as a
nearest point at
comfortable reading
which vision is clear
distance
A diagram may help to clearly establish that we need
do = 25 cm
di = – 100 cm
It is important to remember or realize that the image formed is a virtual image. It is on the
left side of the lens in the diagram above. This means the image distance is negative. Once we
know do and di then it is easy to find the focal length of the lens from the Image Equation,
1= 1 +1
f
do di
1=
1
1
+
f
25 cm – 100 cm
f = 33.3 cm
16. When you look at a color television picture and see a yellow sunflower, the tiny “phospors”
or “color dots” that are lit up, creating this splash of yellow, are
A) cyan and magenta
B) red and blue
C) green and red
D) cyan and blue
17. The image of a real object produced by a concave mirror . . .
A) is always real.
B) may be real or virtual.
C) is always virtual and inverted.
D) is always virtual.
Final - p 10 -
18. The image of a real object produced by a convex mirror . . .
A) is always real.
B) may be real or virtual.
C) is always virtual and inverted.
D) is always virtual.
19. The image of a real object produced by a diverging lens . . .
A) is always real.
B) may be real or virtual.
C) is always virtual and inverted.
D) is always virtual.
20. The image of a real object produced by a converging lens . . .
A) is always real.
B) may be real or virtual.
C) is always virtual and inverted.
D) is always virtual.
21. A real image formed by a single lens is always
A) enlarged and inverted.
B) reduced in size and upright.
C) inverted but may be enlarged or reduced.
D) reduced in size but may be inverted or upright.
22. A virtual image formed by a single lens is always
A) enlarged and upright.
B) reduced in size and inverted.
C) upright but may be enlarged or reduced.
D) enlarged but may be inverted or upright.
23. What is the magnification of a telescope made with an objective lens with a focal length of
50 cm and an eyepiece with a focal length of 2.5 cm?
A telescope’s magnification is given by
f
Mang = θ = – obj
feye
θ’
M = – 50 cm / 2.5 cm
M = – 20
A)
B)
C)
D)
|M| = 20 x
|M| = 25 x
|M| = 30 x
|M| = 50 x
Final - p 11 -
24. Blue light, with λ = 420 nm, shines on an oil film. When viewed under near–normal
incidence, the soap bubble appears very bright. The index of refraction of the oil film is n = 1.2.
What is the thickness of the oil flim at this bright area?
n = 1.0
t
n = 1.2
n = 1.33
A)
B)
C)
D)
125 nm
175 nm
210 nm
420 nm
With air–film–water, there is no “phase shift” so we find a maximum or a bright area when
the total distance traveled, 2t, is one a wavelength,
2t = λ’
2t = (λ/n)
t = (1 / 2) (λn)
t = (1 / 2) (420 nm/1.2)
t =175 nm
25. Calculate the wavelength of the photon that is emitted when a hydrogen atom undergoes a
transition from its n = 3 state to its n = 2 state.
A) 680 nm
B) 656 nm
C) 565 nm
D) 443 nm
1=R 1 – 1
λ
n2
n2
2
1
1=R 1 – 1
λ
22
32
Final - p 12 -
1 = (1.097 x 107 m–1) 1 – 1
λ
4
9
1 = 1.524 x 106 m–1
λ
λ = 6.56 x 10–7 m
λ = 656 x 10–9 m = 656 nm
This wavelength is visible light; it is orange light.
26. A meter stick, with length Lo = 1.00 m when at rest, moves by you at a speed of 0.90 c.
What do you measure its length to be as it speeds by?
A) 0.19 m
B) 0.44 m
C) 2.29 m
D) 5.23 m
This is an example of the Lorentz–Fitzgerald length contraction,
L = Lo SQRT( 1 – v2/ c2 )
L = Lo SQRT ( 1 – 0.92 )
L = Lo SQRT ( 1 – 0.81 )
L = (1.0 m) SQRT ( 0.19 )
L = (1.0 m) ( .44 )
L = 0.44 m
27. A moving stop watch measures a difference in time between two events—the clicks which
start and stop the watch—to be ∆to = 1.00 sect, moves by you at a speed of 0.90 c. What do you
measure for the time between these clicks as it speeds by?
A) 0.19 s
B) 0.44 s
C) 2.29 s
D) 5.23 s
This is an example of the Einstein time dilation,
∆t = ∆to / SQRT( 1 – v2/ c2 )
∆t = ∆to / SQRT ( 1 – 0.92 )
∆t = ∆to / SQRT ( 1 – 0.81 )
∆t = ∆to / SQRT ( 0.19 )
∆t = (1.0 s) / ( .44 )
∆t = 2.29 m
Final - p 13 -
28. Calculate the wavelength of the photon that is emitted when a hydrogen atom undergoes a
transition from its n = 3 state to its n = 2 state.
This is almost one of the homework problems.
1=R 1 – 1
λ
n2
n2
2
1
1=R 1 – 1
λ
22
32
1 = (1.097 x 107 m–1) 1 – 1
λ
4
9
1 = 1.524 x 106 m–1
λ
λ = 6.56 x 10–7 m
λ = 656 x 10–9 m = 656 nm
This wavelength is visible light; it is yellow or orange light.
A) λ = 453 x 10–9 m = 453 nm, blue-green
B) λ = 554 x 10–9 m = 554 nm, yellow
C) λ = 656 x 10–9 m = 656 nm, yellow-orange
D) λ = 703 x 10–9 m = 703 nm, red
29. The “work function” Wo for some metal is 3.7 eV. What is the longest wavelength of light
that released photoelectrons when striking zinc?
This is similar to Example 28.2 from the text.
Wo = 3.7 eV [ 1.6 x 10–19 J / ev ] = 5.92 x 10–19 J
fo = Wo/h = (5.92 x 10–19 J) / (6.626 x 10–34 J s) = 8.93 x 1014 Hz
λ o = c / fo = (3.00 x 108 m/s) / ( 8.93 x 1014 s– 1 )
λo = 3.36 x 10–7 m = 336 nm
A)
B)
C)
D)
λo = 588 nm, yellow
λo = 388 nm, violet
λo = 336 nm, UV
λo = 288 nm, UV
30. Neils Bohr’s Complementarity Principle explains that the nature of both electromagnetic
radiation and matter is
A) wondrously pretty and should be complimented.
B) so complex that both a wave model and a particle model are necessary.
C) so confusing that scientists need to compliment each other to keep their spirits
up.
D) so complicated that only classical electromagnetic theory should be used in
understanding the structure of a hydrogen atom.
Final - p 14 -
31. Heisenberg’s Uncertainty Principle states that
A) life is so uncertain you should always eat dessert first.
B) there are limits to how accurately position and speed of an object may be
known.
C) knowing the momentum of an object better means you will know its energy
less accurately.
D) uncertainties in measuring any quantity are always greater than h, Planck’s
constant.
32. Rutherford’s scattering experiments, involving heavy alpha particles scattered from very
thin gold foil, showed that
A) plum puddings were more popular than fruit cakes in Victorian England.
B) most of the mass and all the positive charge of an atom must be concentrated
in a very small nucleus.
C) the positive charge of an atom was distributed in “lumps” like “plums in a
pudding”.
D) most of the negative charge of an atom was concentrated in a very small
nucleus.
33. Polonium-210 decays by alpha decay in which of these reactions?
Because of the limitations of my word processor, this present notation may look a little “funny”;
146C really should be 14 C
6 .
A) 21084Po —> 20680Hg + 42α
B) 21084Po —> 21084Pb + 42α
C) 21084Po —> 21486Rn + 42α
D) 21084Po —> 20682Pb + 42α
16. While the energy of an atom may have only certain, particular, discrete, quantized values
for the energy,
A) the angular momentum may have any value but its z-component is quantized.
B) the angular momentum and its z-component are also quantized.
C) the angular momentum and its z-component may have any value.
D) the angular momentum is quantized but it may be oriented so that any of its
components may have any value.
18. Carbon-14 decays by beta decay in which of these reactions?
Because of the limitations of my word processor, this present notation may look a little “funny”;
146C really should be 14 C
6 .
Final - p 15 -
A) 146C —> 126C + β –
B) 146C —> 148O + β –
C) 146C —> 147 N + β –
D) 146C —> 168O + β –
33. Thorium-230 decays by alpha decay in which of these reactions?
A) 23090Th —> 22888Ra + 42α
B) 23090Th —> 23492U + 42α
C) 23090Th —> 22886Rn + 42α
D) 23090Th —> 22688Ra + 42α
20. Gamma radiation is
A) high-energy EM waves.
B) long-wavelength EM waves.
C) high-energy hydrogen nuclei ejected by the nucleus.
D) high-energy electrons ejected by the nucleus.
11. Quantum mechanical “tunneling” refers to a particle . . .
A) moving along the magnetic field of a solenoid or tunnel.
B) perpendicular to the magnetic field of a solenoid or tunnel.
C) initially in one allowed region, being found in another allowed region that is
separated from the initial one by a forbidden region.
D) being found inside a forbidden region into which it has tunneled.
12. The atomic
A)
B)
C)
D)
spectrum of gas atoms
involves different wavelengths for absorption and emission.
involves the same wavelengths for absorption and emission.
is continuous for absorption and discrete for emission.
is discrete for absorption and continuous for emission.
Final - p 16 -