fysik 1 kompendium: del 2 standard modellen samt...
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Fysik 1
Kompendium:
Del 2
Standard Modellen
samt Ljus
Klass: Na2
Lärare:
VT14
Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401
FormelSkriv ner här alla formel som du kommer i kontakt med under din läsning. Se till att även skriva ner vad varje variabel betyder och de konstanter som behövs.
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Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401
DefinitionerFyll i några korta meningar som beskriver betydelse av varje term.
De fyra krafterna:
Gravitation:
Elektromagnetiska Kraften
Starka Kraften
Svaga Kraften
nukleon/nuklid
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Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401
ljusets dubbelnatur
Kvantmekanik
LHC/ATLAS
Standardmodellen
kvark
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Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401
Higgs-boson
Frågor
1. Vilka typer av forskning gör dem på Cern?
2. Hur skiljer sig den standardmodellen jämfört med Bohrs eller Rutherfords?
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Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401
3. Vad menas med tomrum? Är det korrekt att säga att en atom mest innehåller luft?
4. Varför kallas det den starka kraften? Beskriv Newtons teoretiska kanon?
5. Vad kom Einstein fram till om ljusets energi?
6. Vad är Unification Teori och varför är det så eftersökt?
7. Ge exempel på ljusets dubbelnatur.
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http://solarsystem.nasa.gov/planets/profile.cfm?Object=Earth&Display=Facts
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Standardmodellen
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* Kvarker påverkas av den starkakraften* Leptons påverkas INTE av starkakraften
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Rekommenderade Övningar: Holt-Physics-Problem-Workbook samt Heurek Kap 13
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Holt Physics Problem Workbook186
NAME ______________________________________ DATE _______________ CLASS ____________________
Copyri
ght ©
Holt, R
inehart
and W
insto
n.A
ll ri
ghts
reserv
ed.
Holt Physics
Problem 23AQUANTUM ENERGY
P R O B L E M
Free-electron lasers can be used to produce a beam of light with variablewavelength. Because the laser can produce light with wavelengths as longas infrared waves or as short as X rays, its potential applications are muchgreater than for a laser that can produce light of only one wavelength.If such a laser produces photons of energies ranging from 1.034 eV to620.6 eV, what are the minimum and the maximum wavelengths corre-sponding to these photons?
S O L U T I O NGiven: E1 = 1.034 eV
= (1.034 eV)11.60 × 10−19
e
J
V2 = 1.65 × 10−19 J
E2 = 620.6 eV
= (620.6 eV)11.60 × 10−19
e
J
V2 = 9.93 × 10−17 J
h = 6.63 × 10−34 J•s
c = 3.00 × 108 m/s
Unknown: lmin = ? lmax = ?
Use the equation for the energy of a quantum of light. Use the relationship
between the frequency and wavelength of electromagnetic waves.
E = hf
f =
l
c
Substitute for f in the first equation, and rearrange to solve for wavelength.
E =
h
l
c
l =
h
E
c
Substitute values into the equation.
lmax =
lmax = 1.21 × 10−6 m
lmax =
lmin =
lmin = 2.00 × 10−9 m
lmin = 2.00 nm
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
(9.93 × 10−17 J)
1210 nm
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
(1.65 × 10−19 J)
15
Problem 23A 187
NAME ______________________________________ DATE _______________ CLASS ____________________C
opyri
ght ©
Holt, R
inehart
and W
insto
n.A
ll ri
ghts
reserv
ed.
ADDITIONAL PRACTICE
1. In 1974, IBM researchers announced that X rays with energies of
1.29 × 10−15 J had been guided through a “light pipe” similar to optic
fibers used for visible and near-infrared light. Calculate the wavelength
of one of these X-ray photons.
2. Some strains of Mycoplasma are the smallest living organisms. The wave-
length of a photon with 6.6 × 10−19 J of energy is equal to the length of
one Mycoplasma. What is that wavelength?
3. Of the various types of light emitted by objects in space, the radio signals
emitted by cold hydrogen atoms in regions of space that are located be-
tween stars are among the most common and important. These signals
occur when the “spin” angular momentum of an electron in a hydrogen
atom changes orientation with respect to the “spin” angular momentum of
the atom’s proton. The energy of this transition is equal to a fraction of an
electron-volt, and the photon emitted has a very low frequency. Given that
the energy of these radio signals is 5.92 × 10−6 eV, calculate the wavelength
of the photons.
4. The camera with the fastest shutter speed in the world was built for re-
search with high-power lasers and can expose individual frames of film
with extremely high frequency. If the frequency is the same as that of a
photon with 2.18 × 10−23 J of energy, calculate its magnitude.
5. Wireless “cable” television transmits images using radio-band photons
with energies of around 1.85 × 10−23 J. Find the frequency of these
photons.
6. In physics, the basic units of measurement are based on fundamental phys-
ical phenomena. For example, one second is defined by a certain transition
in a cesium atom that has a frequency of exactly 9 192 631 770 s−1. Find
the energy in electron-volts of a photon that has this frequency. Use
the unrounded values for Planck’s constant (h = 6.626 0755 × 10−34 J•s)
and for the conversion factor between joules and electron volts (1 eV =
1.602 117 33 × 10−19 J).
7. Consider an electromagnetic wave that has a wavelength equal to 92 cm,
a length that corresponds to the longest ear of corn grown to date. What
is the frequency corresponding to this wavelength? What is its photon
energy? Express the answer in joules and in electron-volts.
8. The slowest machine in the world, built for testing stress corrosion, can
be controlled to operate at speeds as low as 1.80 × 10−17 m/s. Find the
distance traveled at this speed in 1.00 year. Calculate the energy of the
photon with a wavelength equal to this distance.
16
II
Copyri
ght ©
Holt, R
inehart
and W
insto
n.A
ll ri
ghts
reserv
ed.
Section Two—Problem Workbook Solutions II Ch. 23–1
1. E = 1.29 × 10−15 J
C = 3.00 × 108 m/s
h = 6.63 × 10−34 J•s
l =
h
E
c =
l = 1.54 × 10−10 m = 0.154 nm
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
1.29 × 10−15 J
Additional Practice 23A
Givens Solutions
2. E = 6.6 × 10−19 J
C = 3.00 × 108 m/s
h = 6.63 × 10−34 J•s
l =
h
E
c =
l = 3.0 × 10−7 m
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
6.6 × 10−19 J
3. E = 5.92 × 10−6 eV
C = 3.00 × 108 m/s
h = 6.63 × 10−34 J•s
l =
h
E
c =
l = 0.210 m
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
(5.92 × 10−6 eV)(1.60 × 10−19 J/eV)
23Chapter
4. E = 2.18 × 10−23 J
h = 6.63 × 10−34 J •s
E = hf
f =
E
h
f =
6
2
.6
.1
3
8
×
×
1
1
0
0−
−
34
23
J
J
•s
f = 3.29 × 1010 Hz
Atomic Physics
5. E = 1.85 × 10−23 J
h = 6.63 × 10−34 J •s
f =
E
h =
6
1
.6
.8
3
5
×
×
1
1
0
0−
−
3
2
4
3
J/
J
s = 2.79 × 1010 Hz
6. f = 9 192 631 770 s−1
h = 6.626 0755 × 10−34 J •s
1 eV = 1.602 117 33 × 10−19 J
E = hf
E =
E = 3.801 9108 × 10−5 eV
(6.626 0755 × 10−34 J •s)(9 192 631 770 s−1)
1.602 117 33 × 10−19 J/eV
7. l = 92 cm = 92 × 10−2 m
c = 3.00 × 108 m/s
h = 6.63 × 10−34 J •s
h = 4.14 × 10−15 eV •s
f =
l
c
f =
3.
9
0
2
0
×
×
1
1
0
0−
8
2
m
m
/s
f =
E = hf
3.3 × 108 Hz = 330 MHz
17
II
Copyri
ght ©
Holt, R
inehart
and W
insto
n.A
ll ri
ghts
reserv
ed.
Holt Physics Solution ManualII Ch. 23–2
Givens Solutions
8. v = 1.80 × 10−17 m/s
∆t = 1.00 year
l = ∆x
c = 3.00 × 108 m/s
h = 6.63 × 10−34 J •s
∆x = v∆t
∆x = (1.80 × 10−17 m/s)(1.00 year)1365
1
.2
y
5
ea
d
r
ays211
24
da
h
y2136
1
0
h
0 s2
∆x =
E = hf =
h
l
c =
∆
h
x
c
E =
E = 3.50 × 10−16 J
(6.63 × 10−34 J •s)(3.00 × 108 m/s)
5.68 × 10−10 m
5.68 × 10−10 m
E = (6.63 × 10−34 J •s)(3.3 × 108 Hz)
E =
E = (4.14 × 10−15 eV •s)(3.3 × 108 Hz)
E = 1.4 × 10−6 eV
2.2 × 10−25 J
Additional Practice 23B
1. hft = 4.5 eV
KEmax = 3.8 eV
h = 4.14 × 10−15 eV•s
f = }
[KEma
hx + hft]} =
4
[3
.1
.8
4
e
×
V
10
+
−
41
.5
5
e
e
V
V
•
]
s = 2.0 × 1015 Hz
2. hft = 4.3 eV
KEmax = 3.2 eV
h = 4.14 × 10−15 eV •s
KEmax = hf − hft
f =
KEma
h
x + hft
f =
4.
3
1
.
4
2
×
eV
10
+
−
415
.3
e
e
V
V
•s
f = 1.8 × 1015 Hz
3. hft ,Cs = 2.14 eV
hft,Se = 5.9 eV
h = 4.14 × 10−15 eV •s
c = 3.00 × 108 m/s
KEmax = 0.0 eV for bothcases
a. KEmax = hf − hft = 0.0 eV =
h
l
c − hft
l =
h
h
f
c
t
lCs =
hf
h
t,
c
Cs
=
lCs =
b. lSe =
hf
h
t,
c
Se
=
lSe = 2.1 × 10−7 m = 2.1 × 102 nm
(4.14 × 10−15 eV •s)(3.00 × 108 m/s)
5.9 eV
5.80 × 10−7 m = 5.80 × 102 nm
(4.14 × 10−15 eV •s)(3.00 × 108 m/s)
2.14 eV
18