Physics end Flashcards

1
Q

Define and give units for:

electrostatic force Electric potential differenceElectronvolt

A

The force between two charged bodies.is the work done per unit charge on a small positive test charge moved between the points; volts = one joule per coulomb. also called voltagedescribes the energy carried by a particle. The work done when an electron moves through a potential difference of 1 volt; eV

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2
Q

Describe:Hollow conductors:Pear shaped conductor:Corona Discharge:

A

hollow charged objects, within the hollow part there is a zero charge. Concentrated charge density on point because of the difference in size.The strong electric fields, near a point of a pear shape, may be large enough to ionise the molecules in the air in the vicinity of the sharp points. This results in a charge movement away from the conductor

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3
Q

Define and give units for:Electric field strength

A

The electric force per unit charge (q) experienced by a small positive test charge when placed at that point in the field; Newtons per Coulomb or Volts per meter

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4
Q

State:my boi Coulomb’s law

A

The electric force of attraction or repulsion between two charged objects is directly proportional to the products of the two charges and inversely proportional to the square of the distance between their centres.The electric force acts along the line joining the centres of the charges as though the charge were concentrated at a point.

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5
Q

Describe and show units for:Electric current:

A

The flow of positive charge, opposite to the flow of electrons.Measured in the Ampere, or a flow of one coulomb per second.

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6
Q

Derive:E=F/q

A

From coulombs law:F = 1/(4pieo) x (q1q2)/(r^2)E= 1/(4pieo) x q/(r^2)r^2=1/(4pieo) x (q1q2)/FTherefore:E= 1/(4pieo) x q/1/(4pieo) x (q1q2)/FCancels probably to get:E=F/q

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7
Q

Derive:E=V/d

A

Since V=W/q therefore W=VqAnd W=fdAnd F=EqVq=fdVq=EqdV=EdE=V/d

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8
Q

Explain:The electric forces are consistent with newtons laws

A

The magnitude of force acting upon two charges due to their electric fields is the same. Newtons third law states that for every action there is an equal and opposite reaction. Electric forces are consistent with newtons third law.

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9
Q

Describe:The motion of charged particles parallel or antiparticle to a uniform electric fieldComparing the motion of a projectile to motion of a charged particle in a uniform electric field.

A

similar to gravitational acceleration, instead of acceleration due to gravity, there is acceleration due to electric field.

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10
Q

You doing a graph, what you do.What are the units for the lineWhat are the axis

A

HEADINGY OVER XX IS THE INDEPENDANT

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11
Q

Convert energy from electonvolts to joules.Convert energy from joules to electonvolts.Apply the equivalent of W=q♤V

A

Multiply electronvolts 1.6x10^-19 the charge of an electron.Divide joules by 1.6x10^-19 the charge of an electron.W = q♤V = 1.6x10^-19 x 1 = 1.6x10^-19 joules

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12
Q

Define and give units for:Magnetic field.Visual appearance of magnetic fields.

A

The region surrounding a magnetic material or a charge in motion. Such as a current carrying-conductor.This is a vector quantity, appearing as concentric circles around the conductor. The number of field lines indicate the strength of the field at a point.The unit is the Tesla (T) with the symbol B

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13
Q

Describe the right hand rule:

A

The thumb points in the direction of conventional current ant the curl of the fingers indicate the direction of the magnetic field produced.

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14
Q

Describe:Visual representation of current with “x”Visual representation of current with “o”

A

The x refers to a current of magnetic field traveling into the page, and the dot refers to the current or magnetic field travelling out of the page.

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15
Q

Describe:SolenoidsFactors affecting strength of magnetic field::

A

Many coils of wire. Acts like a bar magnet, having a north and south pole.This is also referred to an electromagnet. The magnetic field inside the loops are stronger than outside.Factors that may change the magnetic field strength are:The number of coilsThe magnitude of current traveling throughPresence of an iron or conductive core

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16
Q

Express: magnetic field strength, in relation to gravitational and electric field strength-

A

Magnitude is directly proportional to the current. As gravitational field strength is proportional to the massElectric field is directly proportional to charge. Magnetic field strength is inversely proportional to the radial distance from the conductor. Gravitational field strength is inversely proportional to the distance from a mass.Electric field strength is inversely proportional to the distance from a charge.B=F/il

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17
Q

describe and illustrate:Motion of magnets in magnetic fields-

A

Anything that produces a magnetic field will experience a force within a magnetic field, like a compass which is essentially a small magnet.

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18
Q

Express-Force on a straight current carrying conductor is Proportional to:And On a charged particle

A

Current flowing through the conductorLength of conductorMagnetic field strengthSin/Cos of angle to magnetic fieldOn a charged particle, it is dependant on the magnitude and direction of the velocity.

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19
Q

Describe-Direction of the force on a current-carrying conductor and individual charges:

A

Using the right palm rule The thumb points in the direction of conventional current and the fingers point in the direction of magnetic field. The force acts perpendicularly away from the palm of the hand. A positive charge will travel like this, a negative charge will travel in the opposite direction.The force on a current element that is parallel or antiparallel to a magnetic field is zero.

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20
Q

Describe-Motion of charged particle in uniform magnetic field:

A

Since a charged particle, moving at right angles to a uniform magnetic field, experiences a force of constant magnitude at right angles to the velocity. The charge changes direction without a change in speed. The magnetic force provides the centripetal acceleration required for uniform circular motion.

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21
Q

Derive-Radius of the circular path of a charged particle moving at right angles to a uniform magnetic field of magnitude B.

A

F(centripetal)=F(magnetic)(MV^2)/r=qvBr=MV/qB

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22
Q

Describe-Cyclotron components:

A

An electromagnet is positioned above and below two D-shaped hollow conductors, Producing a uniform magnetic field inside the dees.

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23
Q

Derive and explain meaning of-Period of circular motion of an ion in a cyclotron:

A

T=(2piR)/vBut r=(mv)/qBT=(2pim)/qBThe period does not depend on speed of the ions. Meaning the period of all ions in cyclotron are the same, meaning they reach the gap at the same time interval, regardless of the speed, thus the potential difference can be reversed on a regular basis.

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24
Q

Derive and explain meaning of-Kinetic energy of ions emerging from cyclotron:

A

Ek=(Mv^2)1/2 But r=mv/qB, so v=(rqB)/mK=(M(rqB/m)^2)½K=(rqB)^2/(2m)The kinetic energy is independent of the potential difference across the dees. A larger potential difference does not mean more kinetic energy to the ions, it means the ions will cross the electric field fewer times before they emerge from the cyclotron.

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25
Q

Describe and express with units:magnetic flux-

A

Is a measure of the number magnetic field lines passing through an area. Being the product of magnetic field strength B and the area A. Flux=BAcos(theta)Where theta is the angles between the area and a line normal to the magnetic field.The units for magnetic flux is the Weber (Wb)

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26
Q

ExpressUnits of area conversion, mm, cm, m, km

A

cm^2 -> x10^-4 -> m^2mm^2-> x10^ -6 -> m^2mm^2 -> x10^-2 -> cm^2km^2 -> x10^-6 -> m^2

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27
Q

DescribeElectromagnetic induction:

A

A changing magnetic flux induces a potential difference in a conductor.This induced emf (electromotive force) causes a current to flow. Electrons in the conductor are forced to move given by F=qvB sin(theta). This causes one end of the conductor to become more negative or positive.

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28
Q

Describe and ExpressFaraday’s law:

A

The induced emf is equal to the rate of change of the magnetic flux.e=Change in flux / change in timeThe SI unit of emf is the volt (V)This may be continued to a situation where there are several loops, giving the equation:emf=(n) Change in flux / change in time.

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29
Q

Express:Emf for a changing area or magnetic field:

A

For a changing areaemf= (n) (B) (A2-A1)/ change in timeFor a changing magnitude of magnetic field strength.emf=(n) (A) (B2-B1)/ Change in time

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30
Q

Describe:Lenz’s lawapplication of lenz’s law to solenoid

A

Lenz’s Law states that the induced emf creates a current in a direction that opposes the change in magnetic flux. This may occur through the motion of a magnet and a solenoid, causing the poles to reverse depending on the motion of a magnet, to repel or attract the magnet in motion.

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31
Q

Describe:Lenz’s law in terms of the conservation of energy

A

If the induced poles of the solenoid were to be opposite, then, the magnetic force of attraction between the two unlike poles would cause the magnet to accelerate towards the solenoid. Inducing a greater current in the solenoid, the magnet would further accelerate. This would continue to speed up, increasing the kinetic energy. The energy created would violate the law of conservation of energy. The induced current must therefore flow in a direction to oppose the change in magnetic flux.

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32
Q

Describe:Eddy current:

A

Is the current induced in little swirls on a large solid conductor due to a changing magnetic field. These occur in accordance with Lenz’s law so that they have a direction that will flow to oppose the change in magnetic flux.

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33
Q

Describe:Generators:

A

A generator uses a fixed magnet to generate emfs in rotating conducting loops for electricity production. A generator therefore converts rotational mechanical energy into electrical energy.

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34
Q

Describe:Lamentation:

A

To avoid large eddy currents being produced in a solid conductor, slits can be cut into the metal.

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35
Q

Describe and explain:Transformers:

A

Allow generated voltages to be either increased or decreased before they are used. This is done through primary and secondary coils, each coil with it’s own number of loops. Given the relationship:VpNs=VsNpThese are wrapped around an iron core to increase the strength of the magnetic field. Since the same magnetic flux is passing through both coils, the emf for both are the same. Given that the number of loops vary, then the potential difference will change, changing the volts.Though energy is conserved, so it is transferred from current.

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36
Q

Describe:Step-up and step-down transformers.

A

A step up transformer.When the number of coils in the secondary coil is greater than the primary. Resulting in a greater potential difference in the secondary coil, increasing voltage. The output current is lower than the input currentA step down transformer.When the number of coils in the secondary coil is lesser than the primary. Resulting in a lesser potential difference in the secondary coil, decreasing voltage. The output current is greater than the input current.

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37
Q

ExplainWhy cyclotrons are positioned near the place where they need to be used?What if they were not in the same place?

A

Used to create medical isotopes or in the industry.Isotopes have short half lives, thus they need to be kept in the hospital where they will be used, so that the isotopes do not decay before used. If there was not a readily available cyclotron, then the isotope would have to be made elsewhere and transported. The time would require more of the isotope to be made to compensate for the radioactive decay.Having cyclotron on site reduces costs and increases the number of scans available to patients in time. Some isotopes have short half lives, making them impractical to ship

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38
Q

Describe:Components of generators:

A

Magnets: Produce a uniform magnetic fieldIron core: A soft iron that serves to maximise the magnetic field. It is usually laminated to reduce effect of eddy currents which are bad.Rotator coils: Loops of wire around the iron core. A current is induced in the rotor coils as they rotate in the magnetic field.Split ring communicator: 2 metal half rings connect either end of coils to brushes. Brushes: 2 blocks on either side of communicator. Provide electrical contact to an external circuit.Axle: A rod passing through iron core and coils to provide an axis or rotation for the rotator coils.AC: Induced current is not reversed every half rotation of the coils.DC: Induced current is reversed every half rotation of the coils.

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39
Q

Describe:Why the high voltage is used in transmission lines from generator to consumer.

A

This is to keep the current low, as some energy is lost to the surroundings as heat due to resistance. To minimise the energy loss, the current is kept low by using a high transmission voltage.

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40
Q

you’re given a she question, what you do?

A

Identify the key pointsExplain the points relation to physicsImpact on society

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41
Q

Describe:function of cyclotron

A

Charged particles are accelerated by a uniform electric field in the. The electric field does work on the ions and transfers kinetic energy to the ions (W=qV=Ek=(mV^2)/2).The uniform magnetic field causes the charged particles to move in a circular path so that they return to the electric field by providing centripetal acceleration. The current passing through each dee, alternates, reversing the electric field before the particle returns, accelerating it further.

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42
Q

Explain:transmission of radio or television waves.

A

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.

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43
Q

Explain:Reception of radio or television waves.

A

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)

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44
Q

Describe:orientation of the receiving antenna

A

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

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45
Q

Describe:Electromagnetic waves

A

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

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46
Q

Explain:How EM waves are produced

A

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

47
Q

Describe:Monochromatic lightCoherent wave source

A

is light composed of a single frequency with waves that radiate in all directions away from the sourcewave sources that maintain a constant phase relationship with each other, having the same frequency.

48
Q

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

A

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)

49
Q

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

A

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.

50
Q

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

A

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.

51
Q

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

A

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

52
Q

Describe:Diffraction

A

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

53
Q

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

A

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.

54
Q

Describe:how two-slit interference is produced in the laboratory

A

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

55
Q

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

A

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.

56
Q

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

A

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

57
Q

Explain:Graphing of intensity distribution of diffraction grating

A

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

58
Q

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.

A

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.

59
Q

Derive:dsin(theta)=m(lander)

A

Path difference = dsin(theta) [trig]mLander gives path difference for mth order.

60
Q

Describe:Determining max orders for diffraction grating

A

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

61
Q

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

A

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.

62
Q

Describe:the white-light pattern produced by a grating.

A

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.

63
Q

Explain:How a grating is useful in spectroscopy

A

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

64
Q

Describe:The photoelectric effect

A

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

65
Q

Describe:The work function W

A

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

66
Q

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

A

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.

67
Q

Explain:Maximum kinetic energy of electrons emitted in photoelectric effect

A

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

68
Q

Explain:How x-ray photons can be produced

A

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.

69
Q

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

A

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

70
Q

Convert joules to eV

A

Joules / charge e

71
Q

Explain:Experimental test to find relationship between Ekmax and frequency of light

A

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 = EkmaxW = qDeltaV = eVs = Ekmax

72
Q

Describe:Graph of Ekmax against frequency light.

A

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

73
Q

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

A

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.

74
Q

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

A

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

75
Q

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

A

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

76
Q

Describe:Attenuation of X-rays

A

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

77
Q

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

A

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

78
Q

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

A

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

79
Q

Describe:Minimum exposure time for X-ray photographs

A

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.

80
Q

Describe with equation:De Broglie wavelength

A

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

81
Q

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

A

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

82
Q

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

A

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.

83
Q

Describe:Atomic spectra

A

Range of frequencies of EM radiation emitted/absorbed by matter. 3 types of spectra:Continuous, Line emission (Colour lines), Line absorption (Black lines)

84
Q

Describe:Incandescence

A

All matter vibrates, charges do so and release EM radiation. Increased temperature, increases oscillations of vibrations, thus frequency of EM, which can result in visible light being emitted.

85
Q

Describe:the changes in the spectrum of a filament globe as the temperature of the filament increases.

A

Increase in temperature will cause higher frequency EM radiation to be released, closer to violet colour.

86
Q

Describe:characteristics of the line emission spectra of elements.

A

When atoms of pure gas is heated or subject to potential difference. Light emitted is viewed through spectrometer/diffraction grating, showing color lines on dark.Specific to each gas, thus can identify them.

87
Q

Explain:the uniqueness of the spectra of elements can be used to identify the presence of an element.

A

Emission spectra is specific to each atom as electron shells are different for each. Thus, a known gass’s emission spectra can be used to define its presence via discrete frequencies of light absorbed.

88
Q

Explain:production of characteristic X-rays in an X-ray tube.

A

Produced through incident electrons colliding and removing electrons in lower energy levels of target metals. Higher energy level electrons drop down to fill place, releasing X-ray photon with energy equal to the energy difference between the energy levels.

89
Q

Describe:Energy levels

A

There are different states in an atom, each with their own energy, which can be represented in energy-lvl diagram. Can be raised to excited states through having one electron not in ground state.

90
Q

Explain:magnitude of the transitions on an energy-level diagram relation to the region in the electromagnetic spectrum of the emitted photons

A

From higher to lower, photons are emitted with energy equal to the difference of energy levels the electron passed. Units of energy in eV

91
Q

Describe:Ionisation energy of an atom

A

the minimum energy required to remove a single electron from the atom in its ground state.

92
Q

Describe:line absorption spectrum of atomic hydrogen.

A

Only one electron with series dependant on ending level from 5 up.Lyman series. N = 1 [UV]Balmer series. N = 2 [Visible]Paschen series. N = 3 [IR]

93
Q

Explain:why there are no absorption lines in the visible region for hydrogen at room temperature.

A

At room temperature, electron is in ground state, meaning that any transitions that occur will be in the Lymans series, requiring UV light, not visible.

94
Q

Explain:presence of absorption lines (Fraunhofer lines) in the Sun’s spectrum.

A

Sun produces white light, spanning the visible spectrum. Electrons in atoms of the suns atmosphere absorb photons with energy equal to the gap between lower/higher energy states. These incident photos are removed from the total light, resulting in the dark lines.

95
Q

Describe:Fluorescence

A

process of converting high-energy photons into a larger number of lower-energy photons

96
Q

Describe:Stimulated emission

A

a photon with energy corresponding to a transition from a higher-energy state to a lower-energy state is incident on an atom in the higher state, it can stimulate a transition to the lower state. This results in two identical photons

97
Q

Compareprocess of stimulated emission with that of ordinary (or spontaneous) emission.

A

Stimulated emission occurs when a photon of energy corresponding to a transition of high->low energy states is incident on an atom in the energy state, resulting in an incident photon and emitted photon which have equal energy.Spontanious emission results in only one photon release, is also immediate.

98
Q

Describe:Conditions required for stimulated emission to predominate over absorption.

A

A population inversion is required. By having electrons in a metastable state, excited state with longer lifetime. Hence predominating over spontaneous emission.

99
Q

Explain:function of laser

A

Within a laser, the grain medium absorbs energy, being excited to a metastable state usually. Eventual population inversion achieved and stimulated emission occurs. Mirrors continually reflect photons, one is partially reflective, allowing monochromatic, coherent, unidirectional light to pass.

100
Q

Define:Laser

A

Light Amplification by Stimulated Emission of Radiation.Produces intense beam of coherent light

101
Q

Describe”Population inversion in a set of atoms

A

whenever there are more atoms in a higher-energy state than in a lower-energy state. For practical systems, the higher-energy state must be metastable if a population inversion is to be produced.

102
Q

Describe: useful properties of laser light

A

is coherent and monochromatic, unidirectional and may be of high intensity

103
Q

Explain:Safe handling of lasers

A

Protective eyewear that absorb laser lightProtective gloves reduce burnWarning signs for active laser Electrical safetyFire extinguisher

104
Q

Describe:Standard model

A

Describes particles that form fundamental building blocks of matter and forces that govern their behaviour. Fundamental particles are not made of smaller constituentsQuarks, leptons, gauge bosons

105
Q

Describe: gauge bosons

A

Photon - ElectromagneticW, Z - Weak nuclearGluon - Strong NuclearGraviton - Gravitational

106
Q

Describe:Leptons and their propertiesLepton symbols

A

particles that are not affected by the strong nuclear force. Electron = e(-) = -1Tau = t(-) = -1Muon = U(-) = -1 And the neutrino equivalent:(Lepton)neutrino = V(lepton) = 0

107
Q

Describe:Quarks

A

fractionally charged particles that are affected by all of the fundamental forces, combine to form composite particles and are never directly observed or found in isolation.

108
Q

Describe:how protons, neutrons, and other baryons can be formed from different combinations of quarks.

A

Combinations of quarks for Hadrons which are bayrons or mesons. Baryons are formed by 3 quarks that must be all quark or anti quark.

109
Q

Describe:Lepton numbers

A

Lepton numbers can be one of three types:· electronic lepton number,· muonic lepton number,· tauonic lepton number,The lepton number, regardless of type, for a lepton is 1. All other particles have a lepton number of 0.Lepton neutrinos have their equivalent lepton number

110
Q

Describe:Baryon numbers

A

Quark number is ⅓, anti quark number is -⅓ . All other particles have a baryon number of 0All baryons have number of +1AntiBaryon has number -1

111
Q

Describe:Particle and antiparticle collision

A

they annihilate, releasing energy according to the mass–energy equivalence formula: E=(delta)mc(squared)

112
Q

ExplainUnderstanding of energy of photons released when electron transfer down through electron states

A

No energy higher than the ground state n=1Energy = Upper state - Lower state

113
Q
A