topic 3

Question 1

Explain:

transmission of radio or television waves.

Answer 1

Conductors convert electric current into EM radiation and vice versa by alternating current, causing oscillating electrons which oscillate an electric field, producing oscillating magnetic field perpendicular. Frequency of continually reproducing waves equals that of oscillation of electrons.

Question 2

Explain:

Reception of radio or television waves.

Answer 2

Electrons in receiving antenna is forced to oscillate by electric field of radio wave, they will oscillate in same frequency as electric field. Producing alternating potential difference in receiver.
Force (F=Eq)

Question 3

Describe:

orientation of the receiving antenna

Answer 3

Must be in same plane of polarisation of radio or television waves, or parallel to the producing antenna.

Question 4

Describe:

Electromagnetic waves

Answer 4

are transverse waves made up of mutually perpendicular, oscillating electric and magnetic fields which continually reproduce from eachother.

Question 5

Explain:

How EM waves are produced

Answer 5

Oscillating charges produce electromagnetic waves of the same frequency as the oscillation

Question 6

Describe:
Monochromatic light
Coherent wave source

Answer 6

is light composed of a single frequency with waves that radiate in all directions away from the source

wave sources that maintain a constant phase relationship with each other, having the same frequency.

Question 7

Describe:

what is meant by two wave sources being in phase or out of phase.

Answer 7

Sources are in phase if they emit transverse (light) waves with crests or troughs simultaneously. They are out of phase if they emit waves half a wavelength out of step (lander/2)

Question 8

Explain:

why light from an incandescent source is neither coherent nor monochromatic.

Answer 8

Light produced through heating, causes electrons to vibrate randomly with a range of frequencies. Light released is composed of several colours of light, thus does not maintain constant phase.

Question 9

Describe:

constructive and destructive interference in terms of the principle of superposition.

Answer 9

When two or more electromagnetic waves overlap, the resultant electric and magnetic fields at a point are the vector sum of their separate fields.

When the waves at a point are in phase, ‘constructive interference’ occurs.

Question 10

Describe:

Interference pattern For two monochromatic sources in phase, the waves at a point

Answer 10

constructively interfere when the path difference from the sources to the point is mLander
destructively interfere when the path difference from the sources to the point is (m+½)Lander

Question 11

Describe:

Diffraction

Answer 11

Bending of a wave as it passes through an opening or an obstacle

Question 12

Explain:

maximum and minimum amplitudes For two monochromatic sources in phase through double slit

Answer 12

Will give bright and dark fringes, constructive and destructive interference. Distance between bright is fringe separation (delta Y), distance between slits is d, distance between wall is L, path difference is excess length.

Question 13

Describe:

how two-slit interference is produced in the laboratory

Answer 13

Monochromatic light passes through slits, from source or single slit. Acting as coherent light source. Diffraction occurs, producing wave pattern visible on wall

Question 14

Describe:
how diffraction of the light by the slits in a two-slit interference apparatus allows the light to overlap and hence interfere.

Answer 14

Diffraction causes direction of light to bend and disperse like a wave. Thus resulting in a spread of monochromatic light on wall from 2 sources. May arrive in-phase or not because of path difference where they overlap.

Question 15

Explain:

graphing of the intensity distribution for two-slit interference of monochromatic light.

Answer 15

Wave pattern with central maxima, propagate outwards with lander distance (mlander)

Question 16

Explain:

Graphing of intensity distribution of diffraction grating

Answer 16

Maxima on each order, highest intensity on central maxima/0th order, decreased intensity / larger spread on larger orders.

Question 17

Describe:
how diffraction by the very thin slits in a grating allows the light from the slits to overlap and hence interfere to produce significant intensity maxima at large angles.

Answer 17

Light passes through many slits, center will maxima as it constructively interferes. Many slits means there will always be two which light is ½ lander out of phase, adjacent slits slightly out of phase. Until path difference increases to one wavelength where maximum occurs.

Question 18

Derive:

dsin(theta)=m(lander)

Answer 18

Path difference = dsin(theta) [trig]

mLander gives path difference for mth order.

Question 19

Describe:

Determining max orders for diffraction grating

Answer 19

Angle of 90 substituted for theta gives max m, rounded down.

Question 20

Describe:

how a grating can be used to measure the wavelength of light from a monochromatic source.

Answer 20

Light source goes through grating of known d. Angle of 1st order from 0th order is found by using telescope on turntable with grating in centre.

Question 21

Describe:

the white-light pattern produced by a grating.

Answer 21

White central maxima. Next orders have violet before red as violet ha smaller wavelength, thus produces maxima at smaller angle. Further orders may have violet overlap with red.

Question 22

Explain:

How a grating is useful in spectroscopy

Answer 22

Grating lines are close, meaning maxima viewed at large angles, reducing error in small measurement.
Maxima are thin, reduced error in central judgement

Question 23

Describe:

The photoelectric effect

Answer 23

Emission of electrons from material that is illuminated with light of sufficiently high frequency.
Minimum frequency required is threshold frequency f0

Question 24

Describe:

The work function W

Answer 24

Is the minimum energy required to remove an electron from a surface. This is related to the threshold frequency

Question 25

Describe:

concept of photons and the conservation of energy to explain the experimental observations of the photoelectric effect.

Answer 25

Increased f increases max kinetic energy of emitted electrons E=hf (affected by penetration of material).
The intensity of the incident light affects the number, but not the energy, of emitted electrons.
Below f0, insufficient energy to release least bound surface electrons.

Question 26

Explain:

Maximum kinetic energy of electrons emitted in photoelectric effect

Answer 26

Law of conservation of energy is applied.
Energy of photon = Max energy of electron - energy needed to release least bound electron
hf=Ekmax - W

Question 27

Explain:

How x-ray photons can be produced

Answer 27

produced when electrons that have been accelerated to high speed interact with a target. This interaction causes a sudden change in the direction of motion and speed of the electrons. The decelleration results in kinetic energy converted into EM radiation in the form of X-ray photons.
This process has the german name Bremsstrahlung.

Question 28

Describe:

The main features of the spectrum of X-rays produced through electron interaction.

Answer 28

a continuous range of frequencies (bremsstrahlung) due to the various proximities of the electrons with the nuclei in the target
a maximum frequency given by where is the potential difference across the X-ray tube
high-intensity peaks at particular frequencies

Question 29

Convert joules to eV

Answer 29

Joules / charge e

Question 30

Explain:

Experimental test to find relationship between Ekmax and frequency of light

Answer 30

Photoelectric effect takes place with recorded frequency of lights, electrons travel to anode/detector. Anode charge is increased in negativity until no detection, the potential differnce required is stopping voltage (Vs). Here, work done by electric field = Ekmax

W = qDeltaV = eVs = Ekmax

Question 31

Describe:

Graph of Ekmax against frequency light.

Answer 31

Ekmax = hf - W is related to y = mx + c
Gradient gives plank’s constant (h)
Y intercept is work function
X intercept is threshold frequency

Question 32

Describe:
Purpose of X-ray tube;
filament, target, high-voltage supply, evacuated tube, and a means of cooling the target

Answer 32

High voltage power supply - generates high potential difference between anode and cathode.
Cathode is negative and the Anode is positive.
Filament - wire that is heated by low voltage power supply, emits electrons from the cathode to the anode.
Target - Small metal plate with very high melting point Electrons strike target to produce X-ray photons.
Evacuated Tube - There must be a vacuum (air removed)
Cooling System - Prevents target, anode and cathode from overheating. 99% energy transfer is released as heat.

Question 33

Explain:

the continuous range of frequencies and the maximum frequency in the spectrum of the X-rays.

Answer 33

Shows a left skewed graph with high intensity peaks and a maximum frequency. Area is total number of Xray photons.

Question 34

Derive:
formula for the maximum frequency,
f(max)=[e(delta)V]/h

Answer 34

Accelerating potential (deltaV) works (W) on electrons for kinetic energy. Head on collision results in all Ek becoming Xray energy.
W = Ek = qDeltaV = eDeltaV = hfmax
Fmax = dDeltaV/h

Question 35

Describe:

Attenuation of X-rays

Answer 35

Attenuation is reduction in intensity as they pass through material due to absorption and scattering of X-ray photons.

Question 36

Describe:

Attenuation of X-rays with the types of tissue they pass through

Answer 36

Tissue density; higher density tissue produce higher attenuation than lower density tissues
Tissue thickness; Thicker bones produce higher attenuation than less thick tissue
Elements in tissue; Elements with higher atomic number produce greater attenuation.

Question 37

Describe:

Penetrating power of X-rays relation to its energy and frequency.

Answer 37

High photon energy/frequency means highly penetrating(hard) X-rays. Resultant from a high potential difference across X-ray tube.
Vise-versa for low

Question 38

Describe:

Minimum exposure time for X-ray photographs

Answer 38

Less exposure means clear image as less movement. Is ideal, thus more intense X-ray beam needed. Meaning more photons, produced by larger filament current.

Question 39

Describe with equation:

De Broglie wavelength

Answer 39

Wavelength particles have dependant on their momentum.
Lander = h/p
H = planks constant

Question 40

Describe:

two-slit interference pattern produced by electrons in double-slit experiments.

Answer 40

They behave like light. With sections bright and dark similar to fringes.

Question 41

Describe:

the diffraction of electrons by the surface layers of a crystal lattice was observed.

Answer 41

The Davisson Germer experiment shows that an electron beam reflecting off of crystal lattice would not scatter evenly, instead have peaked intensities at angles. dsin=mlander is used to show that lander=h/p is correct.