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2018 HSC Physics Paper Solutions and Explanations

In this post, we reveal the solutions to the 2018 HSC Physics Exam Paper!

How did you go in the 2018 HSC Physics Exam? Read this post to check the answers and explanations to  the 2018 HSC Physics Exam Paper.


2018 HSC Physics Paper Solutions

This paper was broken into two sections.

Section 1 is based on the core topics and has 2 parts:

  • Part A – 20 Multiple Choice Questions
  • Part B – 10 Short Answer Questions

Section 2 is based on option topics and has five questions. We have included the answers to the most popular option topics:

  • Medical Physics
  • Astrophysics
  • From Quanta to Quarks

Section 1 – 75 Marks

Part A – 20 Marks

 A The gravitational force is towards the centre of the Earth, hence the acceleration is towards the centre of the Earth
 B The graph as shown should be intensity vs wavelength, and higher temperatures emit more radiation.
 A The ISS collides with air molecules in the upper atmosphere (air resistance) and loses momentum.
DThe decrease in rotation speed decreases the back EMF, which increases total EMF and current.
B Force is given by \( F = B I L  sin \theta = 1 \times 2 \times 0.08 \times sin(30) = 0.05 N \)
B The saucepan must be a conductor for currents to be induced. A high AC frequency is necessary for a high change in flux and high induced current.
A The acceleration at the surface of a planet is proportional to the mass, and the inverse of the square of radius; Doubling the mass doubles the acceleration, but doubling the radius decreases the acceleration by a factor of four – these combine for a new ‘g’ that is ½ of the old.
 DLattice U is a p-type, as the dopant provides 3 electrons. There is still a band gap between the conduction and valence band, meaning the corresponding diagram is X. (Note: Lattice T is an n-type semiconductor, and band diagram Y corresponds to a metal.)
A A decreased distance to travel will yield a reduced travel time, hence the astronaut can survive the journey.
C The direction of torque must be constant – otherwise the motor will not work. Hence the direction of force on element WX must reverse every 180 degrees.
D An increased g force is due to an upward acceleration of the object containing a person; case D is the only example where this occurs.
 A The electrons will experience an upward force towards the positive plate.

Using the right hand rule (or the left hand rule for an electron), the magnetic field must be into the page to provide a balancing force downwards.

The magnitude can be calculated by equating

$$qE = qvB$$

$$B = \frac{E}{v} = 0.5 T$$

 B The force on the electron will increase, as F = qvB.

This force is the centripetal force of circular motion; by equation \( qvB = \frac{mv^{2}}{r}\), we can conclude that v is proportional to r.

Hence the radius will increase.

 C The relationship between L and T is \(L = \frac{gT^{2}}{(4\pi^{2})} \), so only C or D show this square relationship.

As g is larger on Earth, a given period will be achieved with a longer length compared to Mars, hence the graph for Earth will be above that of Mars.

 D Michelson attempted to demonstrate a difference in travel times for arms parallels / perpendicular to the direction of earth’s travel due to aether wind.
 A A moving observer will see length contraction for all other objects; hence they will see the tunnel contract. Due to relativity of simultaneity, they will see photo 2 being taken before photo 1, as they are travelling towards the location of the photo-2 event.
 CThe graph shows the work function for zinc is the highest; hence any photon energetic enough to extract electrons from zinc will also extract electrons from potassium.
 B The induced current will oppose any change in flux. Via right hand grip rule, the lower solenoid will generate a downward B field. The solenoid above will thus generate an upward B field to counter this change; these two fields will repel one another and the pointer will move up. The direction of current can be found again from right hand grip rule.
 BAn observer inside the bus will see the the ball move towards the front of the bus. From the perspective of a passenger, the ball has both a forward (due to the \(3 m/s^{2}\) slowdown of the bus) and downward acceleration; this will result in a linear motion! The initial velocity of the ball from the perspective of the passenger is zero.
 CThe needle in X will be forced leftward to oppose the change in flux from the rotating galvanometer. Hence, the generated current must be such that the needle moves to the left; this is a negative current. Tracing the wires from X to Y shows that a negative current coming from X will also yield a negative current through Y, and hence the needle will move to the left.


Part B – 55 Marks

Question 21 ( 4 Marks)

(a)The two forces are equal and opposite, according to Newton’s 3rd Law
(b)Mass on Earth and Moon is 70 kg.

Weight on Earth is w = mg = 70 x 9.8 = 686 N

Weight on the Moon is w = mg = 70 x 1.6 = 112 N

Question 22 (6 marks)

(a)The motion of the magnet will result in regions of the disc experiencing a change in magnetic flux as the poles of the magnet move across them.

This will result in an induced EMF (Faraday’s Law) in the disc and eddy currents.

The eddy currents will produce a magnetic field to oppose the change in the flux (Lenz’s Law) which will result in a force and a torque on the disc.

The disc will rotate in the same direction as the magnet

(b)Graph needs to show a sinusoid with period decreasing and amplitude increasing over the first three seconds

Question 23 (5 marks)

(a)The electrons absorb thermal energy (heat) from the heating filament and escape the metal. The electric field in the electron gun does work on the electrons and increases their kinetic energy. When they strike the screen they stop and their kinetic energy is transferred to the phosphor in the screen, which converts it to heat and light.
(b)Y are solenoids (electromagnets) which produce magnetic fields. The magnetic fields are used to deflect the electron beam and scan it across the screen. There must be two pairs of such coils, one for controlling horizontal deflection and one for controlling vertical deflection.

The image is formed by raster scanning, i.e. by scanning the beam across the screen from pixel to pixel and changing the power of the beam (e.g. the no. of electrons) as the beam hits each pixel to change the intensity of that pixel. Once an image is formed, the scanning repeats to refresh the image

Question 24 (5 marks)

(a)The high voltage wires are attached to the supporting structures using insulators in order to electrically isolate them from the ground. Lightning protection is used: wires placed above the transmission lines, and the supporting towers are metal, well earthed and separated from each other. Current from lightning strikes can safely be conducted to the ground
(b)Wires will attract each other if the current is in the same direction.

In order for there to be zero net force on Y, it must be attracted to X by the same strength as it is attracted to Z, hence the current in X is to the right.

$$F = \frac{kI_{1}I_{2}L}{d}$$

$$\frac{kI_{X}I_{Y}L}{d_{XY}} =\frac{kI_{Y}I_{Z}L}{d_{YZ}} $$

$$\frac{I_{X}}{0.75} =\frac{20}{320} $$

$$I_{X} = 50 A$$


Question 25 (6 marks)

(a) Benefit: No power loss due to heating as resistance is zero.  Limitation: Must be kept below their critical temperature in order to be superconducting (have zero resistance), which is difficult and consumes energy.

(b) The table compares the metal and superconductors.

Cooper pairs experience no resistance. Through their interaction with the lattice one electron loses one phonon of energy and the second electron absorbs one phonon. Hence there is no net energy loss for the Cooper pair.

Metal at room temperatureSuperconducting metal below critical temperature
Conduction bySingle electronsPairs of electrons (Cooper pairs)
MechanismElectrons in the conduction band are mobile and can move through the metal.At low temperatures lattice vibrations reduce. An electron will attract nearby positive ions causing lattice distortion and will lose one phonon of energy. The lattice distortion will increase positive charge density which attracts a second electron that absorbs one phonon of energy. These two electrons are paired through the lattice and form the Cooper pair.
ResistanceElectrons experience resistance.

As electrons move through the lattice they collide with positive ions. These collisions are inelastic on average and the electrons lose energy to the lattice (converted to heat).

Cooper pairs experience no resistance.

Through their interaction with the lattice one electron loses one phonon of energy and the second electron absorbs one phonon. Hence there is no net energy loss for the Cooper pair.

Question 26 (4 marks)

Gravitational FieldElectric Field
Acts onMassCharge
Force it mediatesGravitational force of attractionElectric Force
Definition and unitsForce per unit mass:  N/kg

$$g = \frac{F_{G}}{m}$$

Direction is the direction of force on a mass

Force per unit charge: N/C

$$E = \frac{F_{E}}{q}$$

Direction is the direction of force on a positive charge

Acceleration of objectAll masses in a constant field will accelerate with the same acceleration and in the direction of the fieldAll charges in a constant field will experience an acceleration that depends on their charge-to-mass ratio (q/m). Positive charges accelerate in the direction of the field and negative charges in the opposite direction.
Trajectory in a uniform fieldParabolic trajectory (projectile motion)

Force and acceleration are constant in magnitude and direction.

Component of velocity perpendicular to the field remains constant. Particle accelerates in direction of force.


Question 27 (6 marks)

(a)The camera is viewing the projectile and the ruler at an angle resulting in a parallax error. This will result in a systematic error in the horizontal position of the particle
(b)The horizontal velocity remains constant, the horizontal acceleration is zero. The gradient of the graph gives the velocity:

$$v_{x} = -1.6 m/s $$

The vertical acceleration is constant so the vertical velocity changes at a constant rate.

The initial velocity is 4.2 m/s, the final velocity is -4.8 m/s.

The gradient of the graph gives the acceleration: \( a_{y} =  -9.5 m/s^{2} \)


Question 28 ( marks)

(a)$$\Delta Ep = Ep_{f} – Ep_{i}$$

$$= \frac{-GMm}{r_{f}} – \frac{-GMM}{r_{i}}$$

$$= -GMm [\frac {1}{r_{f}} – \frac{1}{r_{i}}]$$

$$= -6.67 \times 10^{-11} \times  7.35 \times 10^{22} \times 20 [\frac{1}{1740 000+500 000} – \frac{1}{1740000}]$$

$$= 1.257… \times 10^{7} J$$

$$= 1.26 \times 10^{7} J$$ 

(b)If the potential energy increases by \( 1.26 \times 10^{7} J \), the kinetic energy must decrease by the same amount to conserve energy.

\( KE_{Initial} = \frac{1}{2} mv^{2} = \frac{1}{2} \times 20 \times 1200^{2} = 1.44 \times 10^{7}  J \)

\(KE_{Final} = 1.44 \times 10^{7} – 1.26 \times 10^{7} = 1.8 \times 10^{6} J \)

\(\frac{1}{2} mv^{2} = 1.8 \times 10^{6} J \)

\(v = 424 m/s\)


Question 29 (8 marks)

(a)Vacuum photocell:

  • Electrons in the photocathode (A) will each absorb one photon.
  • Assuming the photon energy is higher than the work function (frequency is above the threshold frequency for that metal), the energy absorbed from the photons will allow the electron to overcome the work function and be ejected from the metal.
  • The electric field between A and B will accelerate the electrons through the vacuum in the photocell towards B, resulting in a current through the circuit.

Solar Cells:

  • Combining the n and p-type materials will form a p-n junction.
  • Electrons from the n-type will diffuse into the p-type, causing a loss of charge carriers, and causing the n-type to become negatively charged and the p-type to become positively charge in the junction, resulting in an electric field being formed. This region is called the depletion layer.
  • When light strikes the depletion layer (assuming the photon energy is larger than the bandgap) it can be absorbed by an electron in the valence band, which becomes excited to the conduction band, leaving behind a hole.
  • The electric field will accelerate the hole into the p layer and the electron into the n layer. This charge separation creates a potential difference between the p and n type which results in current flowing from p to n through the circuit
(b) Advance in understanding: understanding of band structure in semiconductors

Our understanding of the valence-conduction band interaction, and particularly the band gap has enabled the development of semiconductor technology. In particular, the relationship between the energy of the band gap and the wavelength of emitted or absorbed photons have prompted the use of semiconductors in light generation (LEDs) and energy generation from light (Solar Panels).

For these inventions to be implemented, however an advance in technology was required: purification of semiconductors and doping to control the band gap of semiconductors. With the ability to engineer a precise, controlled band-gap light emitters and absorbers operating at specific wavelengths could be generated, leading to the wide variety of coloured LEDs and the high efficiency of solar panels, resulting in a substantially increased usage of semiconductors.


Question 30 (6 marks)

An increase in the number of electrical appliances being used in houses means the houses will draw use more energy. This will require them to draw more power (P = E/t) and more current from the secondary coil of step-down transformer T2 (P= VI, and V = 240 V = constant).

The step-down transformer T2 is fixed, so if more power is being drawn from its secondary coil, more power must be supplied to its primary coil in the form of an increase in current. The input voltage must remain fixed since the output voltage is constant at 240 V.

Similar, more current will be drawn from step-up transformer T1, and hence more current must be supplied to it.

Effects on

  1. Generator/power station: Rotates at the same speed to keep voltage and frequency fixed. Will require more torque (mechanics energy input) in order to produce more current and hence more power or energy.
  2. Transformers: Voltages remain constant but currents increase. The higher current Increases the heat produced (Joule heat losses) which means some of the energy is converted to heat and the transformers’ efficiency reduces.
  3. Power lines: The increased current will increase the heat produced meaning the some electrical energy is converted to heat. This also results in a voltage drop across the power line depending on the current and resistance of the power line (V = IR).


2. Section 2 – 25 marks

Question 32 – Medical Physics (25 marks)

(a)(i)  MRI is suitable for use in examining soft tissue with high water content. It is useful for locating tumors and soft tissue disease, and to diagnose Multiple Sclerosis (MS).

(ii) The application of a strong magnetic field changes the spin orientation of hydrogen nuclei in the body. From an initial random state, the nuclei align either parallel or anti-parallel. A slight majority of nuclei are parallel – this constitutes the lower energy configuration.

(b)(i) At a boundary between tissues where there is an acoustic impedance mismatch, a fraction of the ultrasound wave will be reflected, and the remainder of the wave will be transmitted.

Due to the differing impedances of the two materials, both waves B and A will experience some reflection at the fat-kidney interface. The reflection coefficient is a product exclusively of the differences and sums of the impedances, regardless of which is the incident and transmissive medium – there should be no difference between the transmission of wave A and wave B. 

(ii) Impedance of kidney tissue, \( Z = \rho \times v = 1050 \times 1560 = 1.638 \times 10^{6}rayls\)

$$R=\frac{(Z_{kidney} – Z_{fat})^{2}}{(Z_{kidney} + Z_{fat})^{2}}$$

Substitute R = 0.01

$$0.01=\frac{(Z_{kidney} – Z_{fat})^{2}}{(Z_{kidney} + Z_{fat})^{2}}$$

$$0.1=\frac{Z_{kidney} – Z_{fat}}{Z_{kidney} + Z_{fat}}$$

$$0.1(Z_{kidney} + Z_{fat})=Z_{kidney} – Z_{fat}$$


Substitute  \( _{kidney} = 1.638 \times 10^{6} rayls\)

\( Z_{fat} = 1.340 \times 10^{6} rayls \)

(c)Criteria for suitability in diagnosis; emission type, half life and chemical compatibility (toxicity).

  • Emission Type: I-124 is a positron emitter – these particles will rapidly self-annihilate and generate a pair of photons, usually gamma rays. These can be used in positron emission tomography for high-quality medical imaging. In addition, positrons have a low penetration and acceptable ionisation strength.  Hence I-124’s emission is appropriate.
  • Half life: from the data, we can see that iodine-124 has a half life of approximately 100 hours. This is a suitable half-life for a medical imaging isotope; not so long that it stays active in the body and causes prolonged harm, but not so short as to become a logistic obstacle.
  • Chemical compatibility – Iodine is a trace element found in the human body; it is unlikely to have any negative chemical / toxicity effects.
(d)The change in frequency caused by the relative motion between a sound source and an observer is called the Doppler Effect.

The Doppler effect is used  in Doppler sonography and ultrasonography. Acoustic waves reflecting off a moving target will experience a change in frequency proportional to the velocity and direction of the target. Hence this allows the flow of fluids in the target to be measured.

Specific applications include:

  • Detection of blood clots by identifying flow slowdowns / stoppages
  • Detection of abnormal aortic flows
  • Monitoring foetal heart rate
  • Measuring a ‘flow velocity waveform’ – the cycle of fast and slow blood flow as the heart pumps.

X-ray ImageCAT Image
Mode of DetectionX-Ray TransmittanceX-Ray Transmittance
Image GenerationX-Ray source and Detector are static.

Patient stands between x-ray source and detector. Density fluctuations in patient’s body show up as intensity variations on photographic film.

X-Ray source and Detector move around a patient 

Patient lies in centre of CT machine, which rotates around patient and takes several X-ray images at different angles. These exposures are computed together into a 3D image of the target structure.

Image QualityLower resolution, 2 Dimensional. Worse contrast than CTHigher Resolution, 3 Dimensional. Better contrast than X-Ray
Patient RiskMinimalHigher than X-ray due to high radiation dose as a great number of x-ray images are taken.
Application It is useful for examining hard tissues such as bone structure (e.g. viewing bone fractures)It is useful for examining and differentiating soft tissue structures with similar densities such as:

  • Viewing deep soft tissue structure such as vertebrae in the spine or brain tumours.
  • Viewing dilatation of the large blood vessels of the heart.
Benefits Quick and cheap to produce images regarding structure of boneCT images have much higher contrast than X-ray and can produce cross sections of the patient or 3D images.

CT scans are much better at looking at soft tissue structures



Question 33 – Astrophysics (25 marks)

(a) (i) One parsec is the distance to an arc or chord that subtends an angle of parallax of 1 arc second.

(ii) Trigonometric parallax measurements are limited by the resolution of the telescope:

\(\Delta \theta = \frac{1.22 \lambda}{D}\)

Where D is the aperture diameter, \(\lambda\) is the observed wavelength and \(\Delta \theta\) is the size of the Airy disc.

A parallax can be observed if the shift in apparent position can be resolved according to Rayleigh’s criterion.

Atmospheric transparency is good for only radio and optical wavelengths. Shorter wavelengths (UV) are blocked by the atmosphere so the smaller parallaxes measurable can only be observed from space.

Ground-based optical telescopes at any altitude also contend with atmospheric turbulence that distorts incoming wavefronts and blurs the image. This decreases effective resolution hence limiting parallax measurements making resolution worse than the above equation for \(\Delta \theta.\)


A: Eclipsing binaryB: Cepheid variable (Type I)
Extrinsic variableIntrinsic variable
 Two stars of different luminosity orbit each other and periodically block light from each other. The large dip (primary eclipse) is where the more luminous star is eclipsed. The smaller dip (secondary eclipse) is where the less luminous start is eclipsed. Peak brightness is when both stars are fully unobscured. One star whose atmosphere pulsates in size and temperature and hence luminosity. During contractions temperature is higher and so is the luminosity and apparent magnitude. When expanded the temperature is lower and so is the luminosity, and apparent magnitude is fainter.

(ii) From graph B:

\(T = 7 – 1 = 6\) days
From the T-L graph, absolute magnitude is \(M = -3.2\).

$$M = m – 5 log (\frac{d}{10})$$
$$d = 10^{(\frac{(m-M+5)}{5})}$$
$$\frac{(m-M+5)}{5} = \frac{(16.3 + 3.2 +5)}{5} = 4.9$$
$$d = 10^{4.9}$$
\(d = 79 400 \text{pc}\) (3 sig. fig.)

 (c)Star X starts as main sequence blue giant B-class star with core hydrogen fusion proceeding via the CNO cycle. When sufficient helium is deposited into the core, hydrogen fusion continues in a shell around the core and helium core fusion begins.

Increased radiation pressure expands the outer layers causing cooling and the star is now a red supergiant.

Carbon from the helium fusion can fuse in the core while hydrogen and helium fuse in surrounding shells (outer shells with lighter elements). The core region is like shells of an onion.

Oxygen fusion should occur yielding a larger, redder, red supergiant.

The star ends life as a supernova explosion (creating many other elements) while the core of oxygen, neon and other elements collapses into a neutron star.

Note: A flowchart may also be sufficient as long as it did not compromise on detail.

 (d)Photoelectric technologies (e.g. CCD cameras) have improved measurements in astrometry, photometry and especially spectroscopy.

Higher spatial resolution over photographic plates yields better parallax and distance measurements.

Higher accuracy and reliability in photometry yields improved understanding of how colour relates to other properties of stars (luminosity etc) with wider frequency coverage.

Spectroscopy sees the greatest benefits. Photographic spectra are 2D images where relative spectral line intensities cannot be determined quantitatively accurately. Photoelectric detectors are not just more sensitive to a wider range of wavelengths (hence yielding more chemical information from other spectral lines) but allow 1D spectra to be produced easily where spectral line intensities can be quantified precisely. This improves calculations of redshift (or blueshift) for radial velocities and line broadening due to Doppler broadening (thermal and rotational) and pressure broadening. Across the board, furthermore, CCDs yield faster data collection, reduced data loss and improved data comparability.

 (e)Emission Spectra:

Stellar emission spectra are observed by looking at the atmosphere of the star against the backdrop of space. That means observing hot gas against a colder background. The resulting spectrum shows emission lines of various species (chemical composition) and which lines of which species are seen indicates the temperature of the stellar atmosphere. In a school lab, one can observe a gas lamp (e.g hydrogen or mercury) or fluorescent light tubes through a spectroscope in a darkened room.


Blackbody Spectra:

These are produced by blackbody radiators such as stellar cores. Since such objects are never without a surrounding atmosphere, they are not seen directly but as the underlying envelope of stellar absorption spectra.

Blackbody spectra alone yield no chemical information (is a smooth spectrum by definition) they indicate accurately the surface temperature of the star from the distinctive peak wavelength of the spectrum, where \( \lambda_{peak}\) is proportional to \(\frac{1}{temperature}\). This means higher temperature yields shorter peak wavelength (hotter, bluer stars).

In a school lab, one could observe the glowing tungsten filament of an incandescent light bulb through a spectroscope in a darkened room.


Absorption Spectra:

These are produced by observing relatively cooler gas against a brighter and hotter background. Hence stellar absorption spectra are obtained by observing a star directly.

The result is a blackbody spectrum (from the inner stellar core) superimposed with dark lines or dips. The peak of the blackbody yields temperature while the absorption lines yield chemical composition since all atoms are ions have unique characteristic wavelengths of absorption (and emission). Which line of which species also indicates temperature.

In a school lab, one can observe a white light through a transparent box of gas (e.g. hydrogen, oxygen, air) or by dispersing a sample in a flame and observing the flame with a bright backlight using a spectroscope in a darkened room (e.g. AAS in Chemistry).


Question 34 – From Quanta to Quarks (25 marks)

(a)(i) Bohr’s model could not predict atomic spectra for atoms containing more than one electron. It did not explain the mechanisms of stable orbits. It could not explain the Zeeman effect or hyper fine splitting of spectral lines.

(ii) Bohr’s model proposes that electrons exist in specific, stable orbitals around the nucleus. These orbitals have a specific energy. Electrons at a high energy orbital may decay to a lower energy orbital by emitting a photon of light with an energy equal to the difference of the two orbitals; because all of these orbitals are discrete, specific levels, there are only a few photon energies (hence wavelengths) that will be emitted.

(b)(i) Chadwick used the laws of conservation of momentum and energy to identify the mass of particle X. He measured the final momenta and energy of the ejected protons, and used these conservation laws to infer the mass of the particle X.

(ii) Mass of Reactants: 4.0012+9.0122 = 13.0134 amu.

Mass of Products: 12 + 1.0087 = 13.0087 amu.

Mass defect = Mass of Reactants  –  Mass of Products = 0.0047 amu.

Convert to kg by multiplying by \(1.661 \times 10^{-27} = 7.8067 \times 10^{-30} kg. \)

Convert to joules via \(E = mc^{2} = 7.0260 \times10^{-13} J \)

(c)Graph of kinetic energies of beta particles after a specific beta decay reaction. Graph should be a distribution showing a continuous range of different energies are possible – this contrasts to the ‘expected’ kinetic energy, which should be a single value – this expected value should be shown on the graph, at the highest energy point.

Pauli’s proposal of the neutrino was a solution to the energy distribution of beta particles in beta decay – for a given decay the total energy released was constant, and it was expected that the kinetic energy of the released beta particle would be constant – as it is by far the lighter of the two products, it should receive all of the released energy. This was not found to be the case – the energy of beta particles varied substantially – suggesting a violation of conservation of energy. Pauli suggested an unobserved particle may be responsible, carrying some energy away from the reaction without being detected – the neutrino. As it was undetected until this stage, it must have no charge, low mass and minimal interactions with other particles

(d)From the diagrams, we can see that the decay products of

  • source X are likely alpha particles – their cloud trail is short and wide
  • source Y are likely beta particles – their cloud trails are long and thin

Each nucleus of source X undergoing alpha decay would lose 4 nucleons total; 2 neutrons and 2 protons.

Source Y is undergoing beta decay, most likely beta-minus (as positron decay would produce short tracks due to the positron annihilating itself with an electron).

In each nucleus of Y undergoing this decay, one neutron is converted into a proton, with an electron (beta-minus particle) being emitted along with an anti-neutrino.

(e)The standard model describes what matter is made of and how it interacts. According to the standard model matter is made up of quarks and leptons, and interacts through fundamental forces by exchanging force carrier particles. 

There are six quarks (u, d, c, s, t, b) and six leptons (e, \mu, \tau, and three neutrinos), as well as their antimatter counterparts. The fundamental forces are the strong force (mediated by gluons) that acts on quarks and gluons, the weak force (mediated by W and Z bosons) that acts on quarks and leptons, and the electromagnetic force that acts on charged particles (mediated by photons). (The standard model does not explain gravity.)

Out current understanding of the atom is that it is composed of a nucleus consisting of positively charged protons and neutral neutrons, surrounded by negatively charged electrons in discrete energy levels.  The electrons interact with the nucleus through the electromagnetic force mediated by photons. The electrons are attracted to the nucleus through this force which binds them to it. 

In the nucleus, the protons and neutrons (nucleons) are examples of hadrons – composite particles made of three quarks (uud for protons, and udd for neutrons). The quarks in each hadron are held together by the strong force, mediated by gluons. The nucleons are held together by the (residual) strong force which is attractive at the typical separation between nucleons. In a stable nucleus, the (residual) strong force overcomes the electrostatic repulsion between the protons (due to their positive charge) and is able to bind both protons and neutrons to the nucleus. The neutrons are required as they increase the attraction due to the strong force without increasing repulsion due to the electric force, and hence they contribute to the stability of the nucleus.


Written by Matrix Science Team

The Matrix Science Team are teachers and tutors with a passion for Science and a dedication to seeing Matrix Students achieving their academic goals.


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