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.

2

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

3

Relativitet: Heureka Kapitel 13 (s.282-291) samt Ergo 393-401

ljusets dubbelnatur

Kvantmekanik

LHC/ATLAS

Standardmodellen

kvark

4

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?

5

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.

6

http://solarsystem.nasa.gov/planets/profile.cfm?Object=Earth&Display=Facts

7

Standardmodellen

8

* Kvarker påverkas av den starkakraften* Leptons påverkas INTE av starkakraften

9

10

11

12

13

Rekommenderade Övningar: Holt-Physics-Problem-Workbook samt Heurek Kap 13

14

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