Transcript quiz for all chaptersx

A A 
A 
A
1
1
1
1

2
2
2
2
9  3 27

 2.25m 2
12
12
Two point charges –Q and +2Q are placed distance R apart. Where
should a third charge be placed so that it is in equilibrium.
[a] at a point on the right of +2Q
[b] at a point on the left of charge –Q
[c] between –Q and +2Q
[d] at a point on a line perpendicular to line joining –Q and
+2Q
A 22-gauge cylindrical Nichrome wire has a radius of 0.321
mm. The resistivity of Nichrome is ρ= 1.5x10-6 Ωm. If a
potential of 10 V is maintained across a 1.0 m long Nichrome
wire, what is the current in the wire?
A) 22 mA
V
VA V r 2
I 


B) 280 mA
R
L
L
C) 460 mA
D) 1.5 A
E) 2.2 A
The magnitude of the charge on an electron is
approximately:
A)
1023 C
B)
10–23 C
C)
1019 C
D)
10–19 C
E)
109 C
A 5.0-C charge is 10 m from a –2.0-C charge. The
electrostatic force is on the positive charge is:
A)
9.0 × 108 N toward the negative charge
B)
9.0 × 108 N away from the negative charge
C)
9.0 × 109 N toward the negative charge
D)
9.0 × 109 N away from the negative charge
E)
none of these
Two identical charges, 2.0 m apart, exert forces of magnitude 4.0
N on each other. The value of either charge is:
A)
1.8 × 10–9 C
B)
2.1 × 10–5 C
C)
4.2 × 10–5 C
D)
1.9 × 105 C
E)
3.8 × 105 C
An electron traveling north enters a region where the electric field
is uniform and points north. The electron:
A)
speeds up
B)
slows down
C)
veers east
D)
veers west
E)
continues with the same speed in the same direction
If 500 J of work are required to carry a 40-C charge from one point
to another, the potential difference between these two points is:
A)
12.5 V
B)
20,000 V
C)
0.08 V
D)
depends on the path
E)
none of these
A parallel-plate capacitor has a plate area of 0.2 m2 and a plate
separation of 0.1 mm. To obtain an electric field of 2.0 × 106 V/m
between the plates, the magnitude of the charge on each plate
should be:
A)
8.9 × 10¯7C
B)
1.8 × 10¯6C
C)
3.5 × 10¯6C
D)
7.1 × 10¯6C
E)
1.4 × 10¯5C
A ball is held 50 cm in front of a plane mirror. The distance
between the ball and its image is:
A)
100 cm
B)
150 cm
C)
200 cm
D)
zero
E)
50 cm
An object is 30 cm in front of a converging lens of focal length 10
cm. The image is:
A)
real and larger than the object
B)
real and the same size than the object
C)
real and smaller than the object
D)
virtual and the same size than the object
E)
virtual and smaller than the object
An erect object is 2f in front of a convex lens of focal length
f. The image is:
A)
real, inverted, magnified
B)
real, erect, same size
C)
real, inverted, same size
D)
virtual, inverted, reduced
E)
real, inverted, reduced
A 3-cm high object is in front of a thin lens. The object
distance is 4 cm and the image distance is –8 cm. The
image height is:
A)
0.5 cm
B)
1 cm
C)
1.5 cm
D)
6 cm
E)
24 cm
A convex lens has a focal length f. An object is placed
between infinity and 2f from the lens on its axis. The image
formed is located:
A.
B.
C.
D.
at 2f.
between f and 2f.
at f.
between the lens and f.
A.
Which diagram best
represents image I, which
is formed by placing
object O in front of a
plane mirror?
B.
C.
D.
The diagram below shows an arrow placed in front of
a converging lens.
The lens forms an image of
the arrow that is
A.
B.
C.
D.
real and inverted
real and erect
virtual and inverted
virtual and erect
A parallel-plate capacitor has a capacitance of C. If the area of
the plates is doubled and the distance between the plates is
halved, what is the new capacitance?
C/2
C/4
2C
4C
10) In the figure . If R=11Ω, what is the equivalent resistance
between points a and p? 1
A)26 Ω
B) 31 Ω
C)4.6 Ω
D) 36 Ω
Water flows through a pipe. The diameter
of the pipe at point B is larger than at
point A. Where is the speed of the water
greater?
•
a.
Point A
•
b. Point B
•
c.
Same at both A and B
•
d. Cannot be determined from the
information given.
Oil is flowing at a speed of 1.22 m/s through a pipeline
with a radius of 0.305 m. How much oil flows in 1 day?
The volume rate of flow in an artery supplying the brain is 3.6x10-6 m3/s. If the radius of
the artery is 5.2 mm, determine the average blood speed.
Find the average blood speed if a constriction reduces the radius
of the artery by a factor of 3 (without reducing the flow rate).
Examples
A diverging lens with f = -20 cm
h = 2 cm, s = 30 cm
1
1
1


s
s
f
1
1
1
 
30
s
 20
s  12cm
M 
h
s

h
s
h
 12
M  
 0.4
2
30
h  0.8cm
The image is virtual and upright
A converging lens with f = 10 cm
(a) s = 30 cm
1
1
1


30 s 10
s  15cm
M 
s
15
   0.5
s
30
The image is real and
inverted
(b) s = 10 cm
q
(c) s = 5 cm
1
1
1


5
s
10
s   10cm
The image is at infinity
s  10
M  
2
s
5
The image is virtual and
upright
Review Problem:
A ball is dropped from rest and falls towards the earth
due to gravity. After a distance d, the velocity of the
ball is proportional to
1) d1/2
2) d
3) d2
4) d4
5) I have no idea
46
A particle with a mass of 3.5 g and charge of +0.045 C is released from rest at point
A. The electric field strength is E=1200 V/m.
(a) In which direction will this
F=qE
charge move?
(b) What speed will it have after moving
through a distance of 5.0 cm?
• F = qE
 q=+0.045·106 C > 0, Force parallel to E
 Charge accelerates to left
 Work done on charge = qEd = Change in Kinetic Energy




W = (0.045·106C)(1200 V/m)(0.05m) = 54. ·103V ·C =0.054J
(Kf-Ki) = W
(1/2) mvf2 – 0 = 0.054 J
vf2 = 2 (0.054 J) / ( 3.5 ·103kg)= 30.8 m2/s2
 vf =5.6 m/s
Two test charges are brought separately into
the vicinity of a charge +Q. First test charge
+q is brought to point A a distance r from
+Q. Then +q is removed and test charge
+2q is brought to a point B a distance 2r
from +Q. The electrostatic potential felt by
which test charge is greater:
+Q
+q
r
+Q
1) +q
2) +2q
3) It is the same for both
+2q
2r
48
Two test charges are brought separately into
the vicinity of a charge +Q. First test charge
+q is brought to a point a distance r from +Q.
Then this charge is removed and test charge
-q is brought to the same point. The
electrostatic potential energy of which test
charge is greater:
+Q
-q
+q
r
+Q
1) +q
2) -q
3) It is the same for both
r
49
Two test charges are brought separately into
the vicinity of a charge +Q. First test charge
+q is brought to point A a distance r from
+Q. Then +q is removed and test charge
+2q is brought to a point B a distance 2r
from +Q. The electrostatic potential felt by
which test charge is greater:
+2q
+q
+Q
r
+Q
1) +q
2) +2q
3) It is the same for both
2r
50
Rank in order, from largest to smallest, the potential
energies Ua to Ud of these four pairs of charges.
Each + symbol represents the same amount of charge.
1.
2.
3.
4.
5.
Ua = Ub > Uc = Ud
Ua = Uc > Ub = Ud
Ub = Ud > Ua = Uc
Ud > Ub = Uc > Ua
Ud > Uc > Ub > Ua
51
An electron is accelerated from rest through a potential
difference V. Its final speed is proportional to
1. V
2. V 2
3. V 1/2
4. 1/V
5. 1/V ½
6. You need to know the x dependence of V(x)
52
A proton is released
from rest at point B,
where the potential
is 0 V. Afterward,
the proton
1. moves toward A with an increasing speed.
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
53
Consider a charge q in a uniform field E.
i
q
E
d
f
b
How much work does the electric field do in moving
the charge from i to b ?
7) Zero
1)
2)
3)
qEdcosq
qEdsinq
qEd
4) -qEdcosq
5) -qEdsinq
6) -qEd
8) Impossible to determine
54
Consider a charge q in a uniform field E.
i
q
E
d
f
b
How much work does the electric field do in moving
the charge from b to f ?
7) Zero
1)
2)
3)
qEdcosq
qEdsinq
qEd
4) -qEdcosq
5) -qEdsinq
6) -qEd
8) Impossible to determine
55
The electric potential inside a capacitor
1. is constant.
2. increases linearly from the negative to the positive plate.
3. decreases linearly from the negative to the positive plate.
4. decreases inversely with distance from the negative plate.
5. decreases inversely with the square of the distance from the
negative plate.
56
Using potential differences:
From U = qV (potential energy = chargepotential) we
get U = q V.
If 500 J of work is required to carry a 40 C charge from
one point to another, the potential difference between the
two points is
1. 12.5 V
2. 20,000 V
3. 0.08 V
4. depends upon the path
5. none of these
57
During a lightning discharge, 30 C of charge move
through a potential difference of 1.0108 V in 0.02 s.
The energy released by this lightning bolt is
1. 1.51011 J
U = q V.
9
2. 3.010 J
3. 6.0107 J
4. 3.3106 J
5. 1500 J
58
Example
Suppose that a beam of 0.2-MeV photon is
scattered by the electrons in a carbon target.
What is the wavelength of those photon
scattered through an angle of 90o?
A. 0.00620 nm
B. 0.00863 nm
C. 0.01106 nm
D. 0.00243 nm
E. Non of the above
60
Solution
First calculate the wavelength of a 0.2 MeV photon:
E = hc/l= 1240 eVnm/l = 0.2 MeV
l =1240 nm / 0.2 x 106 = 0.062 nm
From Compton scattering formula, the shift is
l = l’l = le (1 – cos 90 ) = le
Hence, the final wavelength is simply
l’ = l +l = le +l  0.00243nm + 0.062 nm = 0.00863
nm
ANS: B,
61
Example
• X-rays of wavelength 0.2400 nm are Compton
scattered and the scattered beam is observed at
an angle of 60 degree relative to the incident
beam.
• Find (a) the wave length of the scattered x-rays,
(b) the energy of the scattered x-ray photons, (c)
the kinetic energy of the scattered electrons, and
(d) the direction of travel of the scattered
electrons
62
solution
l’= l + le (1 - cosq )
= 0.2400nm+0.00243nm(1–cos60o)
= 0.2412 nm
E’ = hc/l’
= 1240 eVnm /0.2412 nm
= 5141 eV
63
p’g
pg
Initial
photon
Eg
me
E’g < Eg
q
f
K
pe
kinetic energy gained by the scattered electron
= energy transferred by the incident photon during the
scattering:
K.E = hc/l - hc/l’=(5167–5141)eV = 26 eV
64
Example
• (c) A 0.0016-nm photon scatters from a free
electron. For what scattering angle of the
photon do the recoiling electron and the
scattered photon have the same kinetic
energy?
65
Solution
• The energy of the incoming photon is
Ei = hc/l = 0.775 MeV
• Since the outgoing photon and the electron each have
half of this energy in kinetic form,
• Ef = hc/l’ = 0.775 MeV / 2 = 0.388 MeV and
l’ = hc/Ef = 1240 eV nm / 0.388 MeV = 0.0032 nm
• The Compton shift is
l = l’ - l = (0.0032 – 0.0016) nm = 0.0016 nm
• By l = lc (1 – cos q )
•
= (h/mec) (1 – cos q ) 0.0016 nm
•
= 0.00243 nm (1 – cos q )
q = 70o
66
Example
•
•
•
•
•
•
•
Which of the following statement(s) is (are) true?
I. Photoelectric effect arises due to the absorption of
electrons by photons
II. Compton effect arises due to the scattering of photons
by free electrons
III. In the photoelectric effect, only part of the energy of
the incident photon is lost in the process
IV.In the Compton effect, the photon completely
disappears and all of its energy is given to the Compton
electron
A. I,II
B. II,III,IV
C. I, II, III
D. III,IV
Ans: E
[I = false; II = true; III = false; IV = false]
67
• To produce an x-ray quantum energy of 10-15 J
electrons must be accelerated through a potential
difference of about
Solution:
• A. 4 kV
The energy of the x-rays photon comes from the
• B. 6 kV
external accelerating potential,V
• C. 8 kV
El  eV
• D. 9 kV
15


1

10
• E. 10 kV V  El / e  11015 J/e= 
eV/e  6250V
19 
 1.6  10 
• ANS: B
68
•
•
•
•
•
•
•
Which of the following statement(s) is (are) true?
I. g -rays have much shorter wavelength than x-rays
II. The wavelength of x-rays in a x-ray tube can be
controlled by varying the accelerating potential
III. x-rays are electromagnetic waves
IV. x-rays show diffraction pattern when passing through
crystals
A. I,II
B. I,II,III,IV
C. I, II, III
D. III.IV
E. Non of the above
Ans: B
69
ConcepTest 27.2e Photoelectric Effect V
A photocell is illuminated
with light with a frequency
above the cutoff frequency.
The magnitude of the current
produced depends on:
1) wavelength of the light
2) intensity of the light
3) frequency of the light
4) all of the above
5) none of the above
Each photon can only knock out one electron. So to
increase the current, we would have to knock out
more electrons, which means we need more
photons, which means we need a greater intensity!
Changing the frequency or wavelength will change
the energy of each electron, but we are interested in
the number of electrons in this case.
ConcepTest 27.2d Photoelectric Effect IV
A metal surface is struck with 1) more electrons are emitted in a given time
interval
light of l = 400 nm, releasing a
stream of electrons. If the light 2) fewer electrons are emitted in a given time
interval
intensity is increased (without
changing l), what is the result? 3) emitted electrons are more energetic
4) emitted electrons are less energetic
5) none of the above
A higher intensity means a more photons, which in turn means
more electrons. On average, each photon knocks out one
electron.
ConcepTest 27.2c Photoelectric Effect III
A metal surface is struck with
light of l = 400 nm, releasing a
stream of electrons. If the 400
nm light is replaced by l = 300
nm light of the same intensity,
what is the result?
1) more electrons are emitted in a given time
interval
2) fewer electrons are emitted in a given time
interval
3) emitted electrons are more energetic
4) emitted electrons are less energetic
5) none of the above
A reduced wavelength means a higher frequency, which in turn
means a higher energy. So the emitted electrons will be more
energetic, since they are now being hit with higher energy
photons.
Remember that c = f l and that E = h f
ConcepTest 27.2c Photoelectric Effect III
A metal surface is struck with
light of l = 400 nm, releasing a
stream of electrons. If the 400
nm light is replaced by l = 300
nm light of the same intensity,
what is the result?
1) more electrons are emitted in a given time
interval
2) fewer electrons are emitted in a given time
interval
3) emitted electrons are more energetic
4) emitted electrons are less energetic
5) none of the above
A reduced wavelength means a higher frequency, which in turn
means a higher energy. So the emitted electrons will be more
energetic, since they are now being hit with higher energy
photons.
Remember that c = f l and that E = h f
ConcepTest 27.2a Photoelectric Effect I
1) metal A
2) metal B
3) same for both
4) W0 must be zero for one of
the metals
A greater cutoff frequency means a higher energy is
needed to knock out the electron. But this implies that
the work function is greater, since the work function is
defined as the minimum amount of energy needed to
eject an electron.
KE
If the cutoff frequency for light in
the photoelectric effect for metal B
is greater than that of metal A.
Which metal has a greater work
function?
f0
Follow-up: What would you expect to happen to the work function
of a metal if the metal was heated up?
f
ConcepTest 27.2a Photoelectric Effect I
1) metal A
2) metal B
3) same for both
4) W0 must be zero for one of
the metals
A greater cutoff frequency means a higher energy is
needed to knock out the electron. But this implies that
the work function is greater, since the work function is
defined as the minimum amount of energy needed to
eject an electron.
KE
If the cutoff frequency for light in
the photoelectric effect for metal B
is greater than that of metal A.
Which metal has a greater work
function?
f0
Follow-up: What would you expect to happen to the work function
of a metal if the metal was heated up?
f
RADIOACTIVITY
• Types of Radioactive decay:
– Beta emission: Converts neutron into a proton by emission of energetic
1
electron; atomic # increases:
n 1 p  0 e
0
1
1
40
0
E.g. Determine product for following reaction:
19 K 1   ?
– Alpha emission: emits He particle.
E.g. Determine product: 226Ra  ? 42 He
88
1
1
0
– Positron emission: Converts proton to neutron:
1 p  0 n  1 e
E.g. Determine product of 94 Tc  ? 0 e
53
1
• Gamma emission: no change in mass or charge but usually part of
some other decay process.
E.g.
14
14
0
6 C 7 N 1e  g
• Electron capture: electron from electron orbitals captured to convert
proton to neutron.
1
p  0 e 1 n
1
1
E.g. Determine product: 40K  0 e  ?
19
1
0
Problem:
A sample contains 4.5g of 13H (tritium), which decays by a - decay to 23He with a
half-life of 12.26 yr. (a) what is the activity of the sample? (b) About how long
would it take the number of tritium atoms to decrease by a factor of a million?
First, we need to find N. The Avogadro number of tritium
atoms (the number of atoms in 1 mole) have a mass of 3g.
N  NA
Decay constant: l 
m g 
A g 
Avogadro’s number
NA  6.0220451023
 9 1023 nuclei
ln 2
0.69
9 1


1.8

10
s
7
t1/2 12.26 y   3.16 10 s / y 
Activity of l N  1.8 109 s 1  9 1023 nuclei   1.6 1015 decays / s  1.6 1015 Bq
the sample:
N  t   N 0 e  lt
 t
ln  N 0 / N  t  
l

1.8 10
ln 106 
9
s
1
3.16 10 s / y 
7
 240 y