2018 HSC Physics Paper Solutions and Explanations

1

2

3

4

5

6

7

8

9

10

11

12

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

$$qE = qvB$$

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

13

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.

14

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.

15

16

17

18

19

20

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

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

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

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.

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

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

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

$$I_{X} = 50 A$$

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).

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.

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

Direction is the direction of force on a mass

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

Direction is the direction of force on a positive charge

Force and acceleration are constant in magnitude and direction.

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

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

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: \( a_{y} =  -9.5 m/s^{2} \)

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

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

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

$$= 1.257… \times 10^{7} J$$

$$= 1.26 \times 10^{7} J$$ 

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

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

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

\(v = 424 m/s\)

• 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

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.

(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.

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

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

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

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

$$0.9Z_{kidney}=1.1Z_{fat}$$

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

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

• 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.

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.

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

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 \(\Delta \theta.\)

(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.)

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.

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.

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).

(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.

(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 \(E = mc^{2} = 7.0260 \times10^{-13} J \)

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

• 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.

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.