Download 64 Exercise Solutions_e

Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
64 Starlight: Messengers from the
Stars
Checkpoint 1 (p. 104)
(a)
angle
(b)
angle
(c)
distance
3.
(c)
F
(d)
F
(a)
Star R
It is 2.512 times brighter.
(b)
Star S
Since the stars are almost the same distance
from us, we can deduce that S has a larger
absolute magnitude from the fact that it has
a larger apparent magnitude.
Note that 1 pc is defined as the distance between
the Sun and a star whose parallax is 1″. Do not
confuse the parallax angle with the parsec.
2.
3.
Mars is farther from the Earth when the angular
diameter is smaller.
(a)
Yes
(b)
Yes
(c)
No.
1 AU ≈ 1.50 × 1011 m.
p.1
apparent magnitude when they look equally
bright in the sky.
Checkpoint
1.
|
4.
Difference in magnitude = 14.5
The increase in brightness is
100(14.5/5) = 631 000 times
5.
∵(1360)∙(12) = I∙(1.52)
∴ I = 604 W m−2
4.
θ = 6779/55.8 × 106 = 0.000 121 rad
5.
See the table below.
In this case, we do not need to convert the
distances to metres if both of them have the same
unit.
distance
star
parallax
Sirius
0.379″
Altair
0.195″
Dumbbell
Nebula
6.
0.00238″
(in pc)
(in AU)
2.64 pc
5.44 × 105 AU
5.13 pc
420 pc
(in ly)
8.60 ly
1.06 ×
106
AU
16.7 ly
8.66 ×
107
AU
1370 ly
Checkpoint 3 (p. 117)
1.
A blackbody is ‘black’ because it does not reflect
any EM radiation.
2.
θ/2 = 1/10 ⟹ θ = 0.2″
The parallax between the Sun and the star is θ/2
but not θ. Then we have θ/2 (in arc seconds) =
1/(10 pc) by definition.
C
3.
(a)
T
(b)
T
(c)
T
(d)
F
(c) < (b) < (a) < (d)
4.
Checkpoint 2 (p. 109)
1.
(a)
yes; no
(b)
yes; yes
(c)
no; no
The difference between apparent magnitude and
absolute magnitude is that we have taken account
of how far the star is from us when considering
the apparent magnitude. Obviously, the farther the
star, the dimmer it appears in the sky.
2.
(a)
(b)
F
P may be farther away and have a smaller
absolute magnitude than Q such that the
stars still have the same apparent
magnitude.
T
By definition, the stars have the same
Note that a class F star has a higher surface
temperature than a class G star which in turn has a
higher surface temperature than a class K star.
Checkpoint 4 (p. 120)
1.
B
Recall that
2.
(a)
.
T
In daily use, luminosity refers to how bright
something is. But in this context, luminosity
Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
refers to how much EM radiation a star
radiates and not only visible light.
3.
(b)
F
Recall that L = 4πR2∙σT4. Luminosity also
depends on the radius of the star.
(c)
F
L = 4πR2∙σT4 = 4π(2.8 × 109)2∙σ(15 000)4
= 2.828 × 1029 ≈ 2.83 × 1029 W
(a)
(b)
(c)
2.
(a)
T
The Sun is a class G star which must have a
higher temperature than a class K star.
2.
3.
(a)
S
(b)
P and Q
Δλ = 589.543 – 588.997 = 0.546 nm
T
From the H–R diagram on p. 123, we can see
that a class B main-sequence star is larger
than a class G main-sequence star.
∴vr = (9.270 × 10−4)∙(3 × 108) = 2.781 × 105 m s−1
We can also deduce this from L ∝ R2∙T4. For a
class B main-sequence star, its luminosity is
100 times that of a class G main-sequence
star but the temperature is only doubled.
Hence a class B main-sequence is larger.
You can determine that the galaxy is moving away
from the shift in wavelength. Longer observed
wavelength ⇒ red shift ⇒ the body is moving
away.
≈ 2.78 × 105 m s−1
The galaxy is moving away from the Earth.
4.
F
Luminosity depends on the temperature as
well as the size.
(a)
The radial velocity is
(1.2 × 105) cos (180°−150°) = 1.039 × 105
≈ 1.04 × 105 m s−1
(b)
The radial velocity is
(1.2 × 106)∙cos (180°−20°) = −1.128 × 106
≈ −1.13 × 106 m s−1
P: red giant
Compare with Fig. 4.24 on p. 121.
Star Y shows a larger Doppler shift because its
radial velocity has a larger magnitude.
5.
(a)
F
A star shows a Doppler shift when the star is
moving with a radial velocity relative to the
Earth.
(b)
T
In reverse, the star is moving towards the
Earth if there is a blue shift.
(c)
T
Recall that
M, G, A, A
When looking up their spectral classes, you
need to consider the surface temperatures
only.
(i)
S<Q<R<P
(ii)
P<Q<R=S
(iii) S < Q < R < P
Note that L ∝ R2∙T4 ⇒ R2 ∝ L / T4.
P is more than 100 times larger in
luminosity than R but has a
temperature of about 1/3 of that of R.
From the above relation, we can see
that P has a larger radius.
3.
No
Δλ/λ = 0.546/588.997 = 9.270 × 10−4
S: white dwarf
(c)
(c)
R does not show red shift or blue shift because its
radial velocity is zero, i.e. it does not move along
the line joining the star and the Earth.
Q and R: main-sequence star
(b)
Yes (no difference)
R2 = 3.85 × 1026 / (4π(5.67 × 10−8)(6000)4)
= 4.169 × 1017
6.
(a)
P
(b)
7.
S
vr = (−8.3 × 10−5)∙(3 × 108) = − 2.49 × 104 m s−1
Hence the expansion speed is 2.49 × 104 m s−1.
Checkpoint 6 (p. 131)
Checkpoint 7 (p. 138)
1.
1.
Yes (no difference)
.
At P, the gas expands (moves) with a velocity
perpendicular to the line of sight, i.e. the
radial velocity is zero. Therefore, the
fractional change in wavelength is also zero.
∴R = 6.457 × 108 ≈ 6.46 × 108 m
(a)
p.2
The light speed observed is independent of
the relative motion of the observer and the
emitter. It depends on the medium where
light travels.
Checkpoint 5 (p. 124)
1.
(b)
|
(a)
R and S
Active Physics Full Solutions to Textbook Exercises
(b)
P
(c)
Q
The radial velocity is positive when the star is
receding from us, negative when approaching us
and zero when moving in a direction to the line of
sight.
71 Photoelectric Effect
p.3
invisible is pulling the celestial bodies and
hence affecting their velocities. This is an
evidence for the existence of dark matter.
Exercise
Exercise 4.1 (p. 110)
1.
2.
|
D
The answer can be found by expressing the
distances in parsecs(pc):
≈ 5.02 × 1011 m
≈ 1.88 × 1033 kg
3.
A
200 ly
61 pc
B
50 pc
50 pc
C
400 000 AU
1.9 pc
D
0.015″
67 pc
2.
A
The distance in parsecs is 1/0.05″ = 20 pc. The
distance in light years is 20 × 3.26 = 65.2 ly.
3.
A
See the table below. The larger the value of θ, the
larger the apparent diameter.
Checkpoint 8 (p. 142)
1.
(a)
(b)
2.
F
Dark matter is ‘dark’ such that it does not
emit or reflect any radiation.
F
In fact, dark matter is the main composition
of the universe and not only found in the
centres of galaxy.
(c)
T
(d)
F
Dark matter is detected as it exerts
gravitational force on visible matter.
(a)
D
F
By Hubble’s law, v = H∙d. That means, the
receding speed depends on the distance
from us.
T
(c)
F
Since the galaxies are receding from us, their
radial velocities are positive. Their spectra
should be red-shifted instead of blue-shifted.
3.
(a)
(b)
The disk is revolving around its centre. The
positive radial velocity suggests that the side
of the disk is receding from us while the
negative velocity suggests the reverse.
The velocity curve suggests that the velocity
does not decrease as predicted. This can be
explained by assuming that something
θ ≈ D/d
A
1.41 × 105 ly
2.54 × 106 ly
0.0555 rad
B
3470 km
384 000 km
0.00903 rad
C
1.39 × 109 m
1.50 × 1011 ly
0.00927 rad
D
150 ly
18 000 ly
0.000833 rad
4.
C
The stellar parallax is (4 × 0.05″)/2 = 0.1″.
Therefore, the distance is 1/0.1 = 10 pc.
5.
C
The ratio can be found by
6.
(a)
The distance is 7000/3.26 = 2147 pc.
The parallax is 1/2147 ≈ 0.000466″.
(b)
In fact, some galaxies close to us, e.g. the
Andromeda Galaxy, are approaching us.
(b)
d
7.
No.
Since the parallax of the star is smaller than
the uncertainty of the measurement, the
distance to the star cannot be measured
accurately.
The max. apparent diameter is
θ = D/d = 3470/363 000 ≈ 9.56 × 10−3 rad
The min. apparent diameter is
θ = D/d = 3470/406 000 ≈ 8.55 × 10−3 rad
8.
(a)
The apparent diameter is
(b)
The actual size is D = dθ
= (2.81× 106)(0.02036) ≈ 5.72 × 104 ly.
Active Physics Full Solutions to Textbook Exercises
9.
71 Photoelectric Effect
(a)
The distance to the nearest star is
(b)
The largest and smallest apparent
magnitudes of the star are 3.4 and 2.1,
respectively.
(b)
The percentage uncertainty is
0.001/0.042 × 100% ≈ 2.38%.
(c)
The difference in magnitude is 3.4 − 2.1 =
1.3.
(c)
Let x be the angular measurement with a
maximum uncertainty of 20%.
|
p.4
Therefore, the brightness of the star changes
by 2.5121.3 ≈ 3.31 times.
Exercise 4.2 (p. 125)
1.
The maximum distance that can be
measured is
Recall the mnemonic ‘Oh Be A Fine Girl Kiss Me’.
d = 1/p = 1/0.005″ = 200 pc.
10.
(a)
X appears brighter because its apparent
magnitude is smaller.
2.
Thus, X is 2.5122.5 ≈ 10.0 times brighter than
Y.
12.
(b)
Both X and Y are at the same distance from
the Earth but X appears brighter. Therefore,
X emits more light per unit time than Y does.
(a)
P and Q appear equally bright as they have
the same apparent magnitude.
3.
C
Since the peaks are the same, the stars should
have the same temperature. As X has a higher peak
for intensity, we can deduce that the star is
brighter due to a larger size (radius)
4.
B
From the H–R diagram, we can see that white
dwarfs have a lower luminosity than
main-sequence stars. In general, the higher the
class of the main-sequence stars, the higher the
luminosity. Therefore, a class B main-sequence
star should have the highest luminosity.
(b)
Q emits more light per unit time as it has a
smaller absolute magnitude.
(c)
Q is farther away from the Earth because Q
emits more light per unit time but it appears
to be as bright as P.
(a)
The difference in magnitude of Mars is
1.83 −(−2.91) = 4.74.
5.
C
Refer to the H–R diagram on p. 121.
Therefore, the apparent brightness of Mars
changes by 2.5124.74 ≈ 78.7 times.
6.
(a)
Yes.
A hotter star has a blackbody radiation curve
peaked at a shorter wavelength. A blue star
emits more light around the blue range
(shorter wavelength). In contrast, a red star
emits more light around the red range
(longer wavelength). Hence we can deduce
that blue stars are hotter.
(b)
No.
Planets are not blackbodies.
(a)
A>G>M
(b)
The distance between the Earth is the
longest when they are on the opposite side of
the Sun, and the shortest when they are on
the same side as shown (not to scale).
Mars
Earth
Mars
Sun
Earth
Sun
7.
The intensity is inversely proportional to the
square of the distance. As a result, the
apparent magnitude can change so greatly.
13.
C
Since the stars are equally bright in the sky, the
distance ratio can be found by
The difference in magnitude between X and Y
is 4.9 − 2.4 =2.5.
11.
D
The temperature drops from class O to class M.
(a)
The period is about 60 h.
Note the time lapse between the two
troughs.
The temperature drops from class O to class
M.
Recall the mnemonic ‘Oh Be A Fine Girl Kiss
Me’.
Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
(b)
13.
8.
9.
(a)
Vega
Its spectrum peaks at a shorter wavelength.
(b)
Class A
Sun is a class G star. As Vega is hotter than a
class G star, it is likely to be a class A star.
(a)
They should have the same (similar) surface
temperature because they are from the same
class.
(b)
Q should be more luminous while P is less
luminous.
(b)
Since L ∝ R2T4, we have
(a)
X: white dwarf
Y: main-sequence star
(b)
Since L ∝ R2T4, we have
|
The ratio is 4 : 25 = 0.16 : 1.
(c)
Since L ∝ R2T4, we have
(d)
Since L ∝ R2T4, we have
(a)
(i)
P: 10 000 K
Q: 3000 K
(ii)
P: class A
Q: class M
The Sun is a class G main-sequence star. Q is
more luminous because a giant is luminous
than a class G main-sequence star. For P, it is
also a main-sequence star but has a lower
class (M as compared with O). Therefore, it
should be less luminous.
10.
(a)
surface temperature / K
(b)
14.
(c)
See the figure above for the positions.
luminosity: (3) < (2) < (1)
temperature: (2) < (3) < (1)
radius: (3) < (2) < (1)
11.
(a)
(b)
(iii) P: main-sequence star
Q: red giant
By L = 4πR2∙σT4, the luminosity is
4π(3.13 × 1011)2∙(5.67 × 10−8)(3000)4
≈ 5.65 × 1030 W.
(b)
Star Q is bigger.
P and Q have the same luminosity but Q is
colder (lower temperature). From L =
4πR2∙σT4, we can deduce that Q has a larger
radius and hence it is bigger.
(c)
Since L ∝ R2T4, we have
It should be a supergiant.
It has a relatively low temperature and a
large radius as compared with the Sun.
(c)
The distance is (3.13 × 1011)/(1.50 × 1011) ≈
2.09 AU.
The Earth would be engulfed by the star.
12.
(a)
It is a main-sequence star.
The ratio is 0.09 : 1.
p.5
Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
Applying
, the radius is
8
(1.010 × 10 )/(2π) × (1.89 × 104)
= 3.038 × 1011 ≈ 3.04 × 1011 m.
A
Since the body is receding, it should be a red shift
and hence Δλ > 0. Options B and D are incorrect.
By
(b)
, we have
2.
A
It has the largest magnitude of radial velocity.
3.
D
Dark matter does not emit or reflect any EM waves;
it exerts gravitational force on matters and that is
how we can ‘see’ it.
4.
A
Option B is incorrect. The galaxies in the Local
Group are close to us and do not obey Hubble’s
law.
Applying
9.
(a)
, we have
Since the spectral line is red-shifted (Δλ > 0), the
galaxy is receding from the Earth.
(a)
(b)
Applying
, the highest speed is
10.
(a)
C
The periods of A, B and C are TA = 0.25 y, TB =
0.5 y and TC = 1 y, respectively.
(b)
A
The speeds of A, B and C are vA =
2 × 105 m s−1, vB = 105 m s−1 and
vC = 5 × 104 m s−1, respectively.
(c)
8.
(a)
Since
and the values of Tv are the
same for A, B and C, we can deduce that the
stars have the same orbital radius.
Applying
(ii)
TP : TQ = 1 : 2
Since
, we have
11.
12.
(i)
100 km s−1
(ii)
125 km s−1
(b)
The existence of dark matter, which exerts
gravitational force on other matter but does
not emit or reflect any EM radiation, is the
view held by many astronomers as the cause
of the deviation between the theoretical and
observed curves.
(a)
The recession velocity v of a distant galaxy is
proportional to its distance d from us, i.e.
v = H∙d.
(b)
The motion of the galaxies is greatly affected
by the gravity of other galaxies in the Local
Group.
(a)
The distance to the star is
The angular distance in radians is
(a)
vP : vQ = 2 : 1
Therefore, the ratio is 4 : 1.
Applying D = dθ, the distance is d = D/θ =
0.1006/(5.963 × 10−7) ≈ 1.69 × 105 ly.
7.
(i)
(iii) Since
and the values of Tv are
the same for P and Q, we have rP : rQ =
1 : 1.
(b)
6.
The mass is
(c)
Options C and D are incorrect. The recession
velocity obeys Hubble’s law, i.e. v = H∙d.
5.
p.6
The period in seconds is 3.2 × 365.25 × 24 ×
60 × 60 = 1.010 × 108 s.
Exercise 4.3 (p. 143)
1.
|
d = 1/p = 1/0.545″ = 1.8349 ≈ 1.83 pc.
(b)
Applying
, we have
, the orbital speed is
The shift in wavelength is −0.234 nm.
Active Physics Full Solutions to Textbook Exercises
The star is approaching as the observed
wavelength becomes shorter (blue shift).
(c)
71 Photoelectric Effect
15.
(a)
See the diagram below.
|
p.7
The radius in metres is
(25 000)(9.461 × 1015) = 2.365 × 1020 m.
The period is 2πr/v
= 2π(2.365 × 1020)/(220 × 103)
= 6.755 × 1015 s ≈ 2.14 × 108 y.
(b)
The total mass is 1011M☉ = 1.99 × 1041 kg.
(c)
The orbital speed is
(d)
No.
Astronomers believe that there are some
‘invisible’ mass, known as dark matter, that
extend much farther than the visible edge of
the galaxy.
The speed of the star is
(d)
The transverse displacement of the star is
D = vtt = (90 × 103)(365.25 × 24 × 60 × 60)
= 2.840 × 1012 m.
Applying D = dθ, the change in angular
position is
Chapter Exercise
13.
Chapter Exercise: Multiple-choice Questions (p. 149)
(a)
1.
A
Statement (2) is incorrect. An object looks bigger
when it is closer to the observer. That mean, the
parallax becomes larger.
Statement (3) is incorrect. Whether an object is
faint or not depends on its luminosity.
(b)
(c)
The two radial velocity curves are out of
phase because the two stars revolve around
their centres of mass in opposite directions.
(a)
No.
In this case, the radial velocity of the small
star is always zero. No Doppler shift can be
observed on the Earth. Therefore, the
mentioned physical quantities cannot be
determined by spectroscopic analysis.
(b)
D
The larger the value of (M – m), the smaller the
distance. (M: absolute magnitude and m: apparent
magnitude)
3.
A
Statement (2) is incorrect. The luminosity of a
star depends on both the temperature and the
radius.
4.
C
Statement (1) is correct. We can analyse the
chemical composition by observing the absorption
lines on the spectrum.
5.
B
Recall L = 4πR2∙σT4.
6.
C
The small star has a higher orbital speed.
By
, we know that r ∝ v. The smaller
star has a larger orbital radius and hence a
higher orbital speed.
14.
2.
All of them can be measured.
Statement (2) is incorrect. By L = 4πR2∙σT4, Q
should have a larger radius if it is colder than S.
For the orbital radius, it can be found by dθ
where d = 1/p and θ is given.
For the orbital period, it can be measured by
recording the time taken for the small star to
orbit once around the massive star.
For the orbital speed, it can be determined
from the orbital period and the orbital radius
using
.
7.
B
Option D is not the correct answer because it
demonstrates a blue shift.
For the remaining options, the radial velocities
should be vA = 35 km s−1, vB = 58 cos 30°
≈ 50 km s−1 and vC = 0. Since the star in B moves
Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
away with the greatest radial velocity, it should
have the largest red shift.
8.
9.
10.
11.
B
The fractional Doppler shift (Δλ/λ) should be the
same for the spectral lines coming from the same
star.
p.8
is more luminous, given that the stars should have
the same surface temperature.
Chapter Exercise: Structured Questions (p. 151)
17.
(a)
The distance is 1/0.007 41″ = 135.0 pc. (1M)
In light years, it should be 135.0 × 3.26
= 439.9 ≈ 440 ly. (1A)
B
Star Y is receding from the Earth when it is at
position P. The corresponding spectrum should be
red-shifted. In contrast, the star is approaching the
Earth when it is at position R. The corresponding
spectrum should be blue-shifted.
(b)
The angular size in radians is
The actual size is
D = dθ = 439.9 × 0.036 65 ≈ 16.1 ly (1M+1A)
C
Statement (2) is correct. The orbital speed is equal
to the maximum radial velocity. The orbital radius
can be calculated by (2.5 × 104 × 500 × 24 × 60 ×
60)/(2π) = 1.719 × 1011 m.
(c)
L > L☉, R > R☉ and T > T☉. (3A)
Refer to the H–R diagram on p. 123.
(d)
Since L ∝ R2T4, we have
D
For a star of surface temperature around 3000 K,
the blackbody radiation curve peaks in the
infrared range (close to the red end of the visible
spectrum). In contrast, if the surface temperature
is around 20 000 K, the curve peaks in the
ultraviolet region.
(1M+1A)
18.
(a)
By Wien’s displacement law (p. 114), the peak λmax
is given by λmax = b/T where b = 2.90 × 10−3 m K is
a constant and T is the surface temperature in
kelvins.
12.
B
An absorption spectrum is produced when a
spectrum of light passes through a gas and the
absorption lines can be regarded as a ‘fingerprint’
of the gas.
13.
B
Options A and C are incorrect. Since the stars
have the same luminosity, Q must be farther away
so that it looks fainter. By
, the intensity is
inversely proportional to the distance squared.
Therefore, Q’s distance should be 5 times that of P
such that its brightness is 1/52 = 1/25 that of P.
14.
|
A
Note that the line is blue shifted, i.e. Δλ < 0.
Options C and D MUST be incorrect. The fractional
shift is −50/410 = −0.1220. Therefore, the
observed wavelength of the red line should be
656 × (1−0.1220) ≈ 576 nm.
15.
C
The two curves peak at the same wavelength and
this implies that the surface temperatures are the
same.
16.
B
Since L ∝ R2T4, X should be larger in size so that it
W: main-sequence star;
X: main-sequence star;
Y: red giant;
Z: white dwarf
(4A)
[1A for each star’s class and position on the diagram]
(b)
Y should be the farthest. (1A)
It has the highest luminosity but appears the
dimmest as seen from the Earth. (1A)
Z should be the closest. (1A)
It has the lowest luminosity but appears the
brightest as seen from the Earth. (1A)
(c)
Since L ∝ R2T4, we have
(1M+1A)
Active Physics Full Solutions to Textbook Exercises
19.
71 Photoelectric Effect
(a)
The radius and the surface temperature (2A)
(b)
The distance of Arcturus from us is
|
p.9
36.7 × (9.461 × 1015) = 3.472 × 1017 m
By
, the intensity should be
(1M+1A)
(c)
(i)
When d = 10 pc,
(1A)
At t = 0 or 2 years, the spectrum is
red-shifted. This implies that the star is
moving away from the Earth. (1A)
At t = 0.5 and 1.5 years, no Doppler shift has
been observed. This implies that the star is
moving in a direction perpendicular to the
line of sight from the Earth. (1A)
At t = 1 year, the spectrum is blue-shifted.
This implies that the star is moving towards
the Earth at that time. (1A)
(1A)
(ii)
The distance of Arcturus in pc is
36.7/3.26 = 11.26 pc (1M)
If m = −0.04,
(1A)
20.
(a)
(b)
From the graph, the greater the mass, the
smaller the radius and hence the volume. (1A)
Since density is mass over volume, we can
deduce that white dwarfs of larger mass
should have a higher density. (1A)
Since L ∝
R2 T 4 ,
Since L ∝
R2 T 4 ,
(b)
Applying
, we have
(1A)
we have
(c)
The radial velocity of the star is as shown.
(1M+1A)
(c)
we have
(3A)
[1A for axes + 1A for curve + 1A for labels]
(d)
(1M+1A)
(d)
Applying
strength is
The radius of the orbit is
(60 000)(2 × 365.25 × 24 × 60 × 60)/2π
= 6.027 × 1011 m.
Applying
, the gravitational field
, the mass is
(1M+1A)
(e)
(1A)
The weight is mg = (70)(3.67 × 106)
≈ 2.58 × 108 N. (1A)
21.
(a)
Yes, it can be a black hole. (1A)
Its mass is 3.25 × 1031/1.99 × 1030 ≈ 16M☉.
(1A)
22.
(a)
The orbital motion of the star is as shown.
The rotational speeds of the lumps of ice can
be measured by observing the Doppler shift
in the spectra observed from the lumps at
both sides of the rings. (1A)
By
, the radial velocities of the lumps
can be found, and hence the rotational
speeds. (1A)
(b)
The rotational speeds measured would be
differed by a factor of cos 20°. (1A)
Active Physics Full Solutions to Textbook Exercises
71 Photoelectric Effect
The measured speeds are lower than the
actual speeds as all the measurements of
velocities would be differed by this factor. (1A)
(c)
(b)
Different sets of dark lines correspond
to different elements. Hence the
existence of a specific set of dark lines
reveals the existence of an element in
the star. (1A)
(i)
The star is receding and there is a
red-shift for all minima. (2A)
(ii)
Δλ = 120.9 – 119.5 = 1.4 nm (1M)
Applying
(1A)
(d)
Applying
p.10
(ii)
The speed decreases with the distance. (1A)
Suppose the centripetal force on the lumps is
provided by the gravitational force from
Saturn.
|
,
, (1M)
v = (1.4/119.5)∙(3 × 108) (1M)
v ≈ 3.51 × 106 m s−1 (1A)
(1M)
Applying
25.
, the mass is
(a)
It measures how bright the star is as seen
from the Earth. (2A)
(b)
(i)
dY = 3.8 pc ⇒ 1/dY2 = 0.0693 pc−2.
dZ = 4.6 pc ⇒ 1/dZ2 = 0.0473 pc−2.
(1A)
23.
(a)
The recession velocity is directly
proportional to the distance. (1A)
(b)
The relation in (a) suggests that the universe
is expanding. (1A)
This is because the farther away a galaxy, the
faster it is moving away from us. (1A)
(c)
65 km s−1 Mpc−1 (1A)
(ii)
Since v = H∙d, the slope of the graph is
actually the Hubble constant.
(d)
(i)
(ii)
The universe is expanding. (1A)
All distant galaxies, including the
galaxy shown, are receding from us and
thus all the measured spectra are
red-shifted. (1A)
The difference in the degree of Doppler
shifts at the two sides of the galaxies
suggests that the galaxy is rotating. (1A)
Neglecting the recession velocity of the
galaxy, the stars on the right are
receding from us while those on the
left are approaching us. (1A)
Hence the galaxy is rotating in a
anticlockwise direction when viewed
from above. (1A)
(2A)
[1A for data points, 1A for the line]
(iii) k = slope of the graph (1A)
Hence k = 7 × 10−3 W m−2 pc2.
(c)
(i)
log I = −2 log d −2.2 (2A)
(ii)
m = −2.5 (−2 log d −2.2) + a
⇒ m = 5 log d + b
where b = a + 5.5 (2A)
(d)
m – M = 5 log (d/10) (1A)
3.0 – M = 5 log (2.1/10)
⇒ M ≈ 6.39 (1A)
26.
(a)
(i)
Since L ∝R2T4, we have (1A)
(iii) It is the existence of dark matter in the
galaxy that causes the orbital speeds of
the stars increase with their distances
from the galactic centre. (1A)
24.
(a)
(i)
The dark lines are formed when certain
wavelengths of the visible spectrum
are absorbed by the gases on the
surface of the star and then re-radiated
in all directions. (2A)
(1A)
(ii)
From the above,
(1M+1A)
Active Physics Full Solutions to Textbook Exercises
(b)
(i)
By
71 Photoelectric Effect
, the calculated L would be
larger if the measured I remains
unchanged while d becomes larger. (1A)
Since L ∝R2, the radius would also
become larger. (1A)
(ii)
(c)
The parallax is too small. (1A)
The brightness of the sun is
The distance of Betelgeuse in AU is
200 × 206 265 = 4.126 × 107 AU (1A)
The brightness of Betelgeuse is
(1A)
From the above calculation, it is not true that
Betelgeuse is as bright as the Sun. (1A)
27.
(a)
(i)
Substitute L = 4πR2∙σT4 into
,
the intensity is
(1A)
Rearrange I = P/A, the power is
(1A)
(ii)
Assuming that the planet is a
blackbody,
(1A)
Simplify the above and we have
(1A)
(b)
(i)
The temperature is
(1M+1A)
(ii)
Water can be in the liquid form (1A)
because the temperature is between
the ice point and the steam point. (1A)
It may be favourable.
(iii) A class K star is cooler than our Sun. (1A)
This suggests that the surface
temperature can be lower as the power
received from the star is lower. (1A)
|
p.11