ESAT Physics: Waves and radioactivity, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   A wave travels 340 m across the surface of a lake in 2.5 s. If the wave's period is 0.05 s, what is its wavelength?

   1. A 17 m
   2. B 6.8 m
   3. C 2720 m
   4. D 42.5 m
2. 21 mark

   A fishing float bobs up and down on the surface of a pond as ripples pass beneath it, but the float does not drift sideways with the ripples. Which statement best explains this observation, and what does it demonstrate about waves in general?

   1. A The float bobs up and down only because water waves are longitudinal waves, and in a longitudinal wave the particles of the medium do not move at all as the wave passes; only the wave pattern appears to move.
   2. B The float does drift slowly in the direction of the ripples, just too slowly to notice, because a wave must transfer some matter in the direction of travel in order to transfer energy.
   3. C The float bobs up and down because the ripples reflect off the pond's edges and interfere with the original wave, cancelling out any sideways motion; in an infinite pond with no edges, the float would drift with the wave.
   4. D The float bobs up and down only, without drifting, because waves transfer energy from place to place without there being any net movement of the medium itself; each water particle oscillates about a fixed position as the wave passes through.
3. 31 mark

   A cyclist moves towards a stationary emergency siren that emits a constant, unchanging frequency. A second listener stands still, at rest relative to the siren, some distance beyond it. Compared with the frequency the siren actually emits, what frequency does each listener detect?

   1. A The cyclist detects a higher frequency than emitted, because moving towards the source means the cyclist meets each wavefront sooner than if stationary; the stationary listener detects exactly the emitted frequency, since there is no relative motion between the siren and that listener.
   2. B Both listeners detect exactly the emitted frequency, because the siren itself is stationary and only a moving source can create a Doppler shift; the cyclist's own motion has no effect on the frequency detected.
   3. C The cyclist detects a lower frequency than emitted, because approaching the source means the cyclist takes longer to receive each wavefront; the stationary listener detects a higher frequency than emitted, because they are further from the source.
   4. D Both listeners detect a higher frequency than emitted, because the cyclist's motion Doppler-shifts the sound for everyone nearby, including listeners who are themselves stationary.
4. 41 mark

   A plane mirror reflects a fixed ray of light, incident at a constant angle, onto a screen. The mirror is then rotated by 12 degrees about an axis lying in the plane of the mirror, while the incident ray's direction does not change. By what angle does the reflected ray's direction rotate?

   1. A 12 degrees
   2. B 6 degrees
   3. C 24 degrees
   4. D 48 degrees
5. 51 mark

   A ray of light enters a rectangular glass block through one flat face at an angle to the normal, travels through the glass, and exits through the opposite, parallel face back into air. Which statement correctly describes the ray's direction as it exits the block, compared with its original direction before entering the block?

   1. A The ray exits at a completely different angle from its original direction, because refraction at the second surface bends the ray by the same amount, in the same rotational sense, as at the first surface, doubling the total change in direction.
   2. B The ray exits parallel to its original direction, but is laterally displaced (shifted sideways) from where it would have been if it had travelled in a straight line, because the bending towards the normal on entry is exactly reversed by an equal bending away from the normal on exit.
   3. C The ray exits travelling back the way it came, reversed by 180 degrees, because entering a denser medium and then leaving it reverses the direction of travel, similar to reflection.
   4. D The ray exits parallel to its original direction and passes through exactly the same point it would have reached if the glass block had not been there at all, because the two refractions at parallel surfaces cancel out perfectly with no lateral shift.
6. 61 mark

   A ship uses an ultrasound pulse to measure the depth of the seabed beneath it. The pulse is emitted straight down, reflects off the seabed, and its echo is detected back at the ship 0.8 s after the pulse was sent. If the speed of sound in seawater is 1500 m/s, what is the depth of the seabed below the ship?

   1. A 600 m
   2. B 1200 m
   3. C 300 m
   4. D 2400 m
7. 71 mark

   Wave P has twice the amplitude of wave Q, but half the frequency of wave Q. If both are sound waves played through the same speaker, which statement correctly compares how they are heard, in terms of loudness and pitch?

   1. A Wave P sounds louder than wave Q, and wave P has a higher pitch than wave Q, because a larger amplitude always corresponds to a higher frequency.
   2. B Wave P and wave Q sound equally loud, because loudness depends only on frequency, not amplitude; wave P has a lower pitch than wave Q because it has a lower frequency.
   3. C Wave P sounds quieter than wave Q, since a lower frequency wave carries less energy per second and is therefore always quieter regardless of amplitude, and wave P has a lower pitch than wave Q.
   4. D Wave P sounds louder than wave Q, and wave P has a lower pitch than wave Q.
8. 81 mark

   A radio wave has a wavelength of 300 m. A gamma ray has a wavelength of 3 x 10^-11 m. Both travel through a vacuum at the speed of light, c = 3 x 10^8 m/s. What is the gamma ray's frequency divided by the radio wave's frequency, i.e. how many times greater is the gamma ray's frequency?

   1. A 10^-13
   2. B 10^25
   3. C 10^13
   4. D 10^19
9. 91 mark

   An ion has a nucleon number (mass number) of 27, an atomic number of 13, and an overall charge of 3+. How many protons, neutrons and electrons does this ion contain?

   1. A 13 protons, 14 neutrons, 13 electrons
   2. B 13 protons, 14 neutrons, 10 electrons
   3. C 13 protons, 14 neutrons, 3 electrons
   4. D 13 protons, 14 neutrons, 16 electrons
10. 101 mark

    A nucleus of thorium-232 (atomic number 90) decays by emitting an alpha particle. The nucleus produced then decays further by emitting a beta particle. What are the mass number and atomic number of the final nucleus, after both decays?

    1. A Mass number 224, atomic number 86
    2. B Mass number 232, atomic number 89
    3. C Mass number 228, atomic number 87
    4. D Mass number 228, atomic number 89
11. 111 mark

    Alpha, beta and gamma radiation each ionise the atoms they pass close to, but to very different extents, and each penetrates matter to a different depth before being absorbed. Which statement correctly describes the relationship between a radiation type's ionising power and its penetrating power, and correctly ranks the three types?

    1. A Alpha radiation is the most strongly ionising and the least penetrating; gamma radiation is the most penetrating and the least ionising. A type that ionises strongly interacts with, and transfers energy to, the atoms it passes very frequently, so it loses its own energy over a very short distance and cannot travel far into a material.
    2. B Alpha radiation is the most strongly ionising and the most penetrating, because a particle that interacts strongly with matter also travels further before losing its energy; gamma radiation, being the least ionising, is stopped by the thinnest materials.
    3. C Gamma radiation is the most strongly ionising, since it carries the most energy per particle of the three, and it is also the most penetrating, since more energetic radiation always travels further before being absorbed.
    4. D The ionising power and penetrating power of a type of radiation are unrelated to one another; alpha, beta and gamma are ranked in a different order for each property, with beta radiation being both the most ionising and the most penetrating.
12. 121 mark

    A sample of a radioactive isotope has a half-life of 3 days. What fraction of the original number of undecayed nuclei remains after 15 days have passed?

    1. A 1/16
    2. B 5/32
    3. C 1/32
    4. D 1/5
13. 131 mark

    A Geiger counter placed near a radioactive source records a count rate of 144 counts per minute. Background radiation alone, measured with the source removed, contributes 24 counts per minute at all times, and this background contribution does not change over time. The source itself has a half-life of 5 minutes. What total count rate will the Geiger counter record 15 minutes later?

    1. A 18 counts per minute
    2. B 15 counts per minute
    3. C 42 counts per minute
    4. D 39 counts per minute
14. 141 mark

    A nucleus first undergoes alpha decay, leaving the daughter nucleus in an excited (high-energy) state. This excited nucleus then emits a gamma ray to lose its excess energy, without emitting any further particles. What effect does this gamma emission alone have on the nucleus's mass number and atomic number?

    1. A The mass number decreases by 4 and the atomic number decreases by 2, since gamma emission, like the alpha decay before it, removes nucleons from the nucleus.
    2. B Neither the mass number nor the atomic number changes, because a gamma ray is electromagnetic radiation with no mass and no charge; it carries away energy only, not any protons or neutrons.
    3. C The atomic number increases by 1 while the mass number stays the same, because gamma emission, like beta decay, involves a neutron converting into a proton.
    4. D Both the mass number and the atomic number decrease by 1, because the nucleus loses a single nucleon as it settles into a lower energy state.
15. 151 mark

    Water waves travelling across the surface of a swimming pool pass at an angle from a deep region into a noticeably shallower region, and are observed to change direction, bending towards the normal to the boundary between the two depths. This is a standard analogy for the refraction of light at a boundary between two transparent media. Which statement correctly extends the analogy to describe what happens to the water wave's speed, frequency and wavelength as it crosses into the shallower region, and correctly identifies the equivalent property of the optical medium that the shallower water represents?

    1. A The water wave speeds up in the shallower water, its frequency stays the same, and its wavelength increases; the shallower water is analogous to an optically denser medium, in which light also travels more quickly.
    2. B The water wave slows down in the shallower water, its frequency stays the same, and its wavelength increases, because a slower wave takes longer to complete each cycle and so spreads out over a greater distance; the shallower water is analogous to an optically less dense medium.
    3. C The water wave slows down in the shallower water, its frequency stays the same, and its wavelength decreases; the shallower water is analogous to an optically denser medium, in which light also travels more slowly.
    4. D The water wave slows down in the shallower water, its frequency decreases to match the new speed, and its wavelength stays the same; the shallower water is analogous to an optically denser medium, in which light's frequency also decreases on entry.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: B

   1. The wave's speed is distance / time = 340 m / 2.5 s = 136 m/s.
   2. The frequency is the reciprocal of the period: frequency = 1 / period = 1 / 0.05 s = 20 Hz.
   3. Wavelength = speed / frequency = 136 / 20 = 6.8 m.
   4. The answer is B.
   • Why not A: This treats the given 340 m distance as if it were already the wave's speed, skipping the step of computing speed = distance / time = 340 / 2.5 = 136 m/s, and instead multiplies 340 by the period directly (340 x 0.05 = 17).
   • Why not C: This correctly finds the speed as 136 m/s, but then divides speed by the period's numerical value directly (136 / 0.05 = 2720) instead of first inverting the period to find the frequency (frequency = 1 / period = 20 Hz) and dividing by that; it mistakes the period for the frequency.
   • Why not D: This finds a 'speed' by multiplying the given distance and time instead of dividing them (340 x 2.5 = 850 m/s, rather than 340 / 2.5 = 136 m/s), then correctly multiplies this wrong speed by the period (850 x 0.05 = 42.5) to get a wavelength built on the wrong speed.

2. Question 2Answer: D

   1. Waves transfer energy from one place to another without a net transfer of the medium (matter) they travel through.
   2. As a ripple passes beneath the float, the water particles directly under the float (and the float itself) move up and down about a fixed average position, but do not travel along with the wave in the direction it is moving.
   3. This is exactly why the float bobs up and down repeatedly but ends up back close to where it started, rather than being carried across the pond by the ripples.
   4. The answer is D.
   • Why not A: This wrongly claims water waves are longitudinal and that the medium's particles do not move at all; in a longitudinal wave the particles do oscillate, just parallel to the direction of travel rather than perpendicular to it, and in any case some particle motion always occurs as a wave passes. It is the net transfer of matter that is zero, not the motion itself.
   • Why not B: This assumes that transferring energy requires transferring matter in the direction of travel, which is exactly the misconception waves disprove: a wave can transfer energy over a distance while each particle of the medium only oscillates about a fixed average position, with no net drift.
   • Why not C: This misattributes the effect to reflection and interference from the pond's edges; the float would bob up and down without net drift even in a pond too large for edge reflections to matter, because the absence of net matter transport is a general property of wave motion itself, not a result of boundary interference.

3. Question 3Answer: A

   1. The Doppler effect arises from relative motion between a source and an observer, and it does not matter which of the two is doing the moving.
   2. As the cyclist moves towards the stationary siren, they reach each successive wavefront sooner than they would if standing still, so they detect more wavefronts per second than the siren actually emits: a higher frequency.
   3. The second listener has no relative motion with the siren (both are stationary), so they detect exactly the frequency the siren emits, with no shift at all.
   4. The answer is A.
   • Why not B: This assumes only a moving source can produce a Doppler shift, but relative motion between source and observer in either direction produces the effect; a listener moving towards a stationary source encounters wavefronts more frequently than a stationary listener would, and so detects a higher frequency, exactly as if the source had approached them instead.
   • Why not C: This gets the direction of the shift for the moving cyclist backwards: approaching a source means encountering successive wavefronts sooner, not later, which raises the detected frequency, not lowers it. It also wrongly attributes a frequency shift to the stationary listener's distance from the source; being further away changes the wave's amplitude (loudness), not its frequency, and this listener has no relative motion with the siren so detects exactly the emitted frequency.
   • Why not D: This treats the Doppler shift as if it radiated outward from the moving cyclist to affect everyone nearby, but the shift is specific to the relative motion between the source and each individual observer; the stationary listener has no relative motion with the stationary siren, so they detect the true emitted frequency regardless of what the cyclist is doing elsewhere.

4. Question 4Answer: C

   1. When a mirror rotates by an angle through an axis in its own plane, the normal to the mirror (always perpendicular to the mirror surface) also rotates by that same angle.
   2. For a ray incident in a fixed direction, the angle it makes with the new normal changes by 12 degrees compared with before the mirror rotated.
   3. By the law of reflection, the angle of reflection (measured from that same new normal, on the other side) must also change by 12 degrees, so the reflected ray's direction has shifted by 12 degrees relative to the incident-ray side and by another 12 degrees on the reflected-ray side of the geometry, giving a total change of 12 + 12 = 24 degrees in the reflected ray's absolute direction.
   4. The answer is C.
   • Why not A: This assumes the reflected ray rotates by the same angle as the mirror itself. In fact both the angle the incident ray makes with the (now rotated) normal, and the equal angle of reflection on the other side of that normal, shift by the mirror's rotation angle, so the ray's total change in direction is double the mirror's rotation, not equal to it.
   • Why not B: This halves the mirror's rotation instead of doubling it, as if the law of reflection reduced the effect of the mirror turning rather than doubling it; the normal rotates by the full 12 degrees the mirror rotates, and this shift is applied on both the incidence and reflection sides of the ray's path.
   • Why not D: This doubles the effect twice: once for the change in the angle of incidence and again for the angle of reflection, as though these were two separate factors of two to apply. In fact a single doubling (2 x 12 = 24) already accounts for both, since the incidence and reflection sides shift by the same 12 degrees each, adding to one total change of 24 degrees, not 48.

5. Question 5Answer: B

   1. At the first surface, the ray crosses from air (less dense) into glass (more dense), so it slows down and bends towards the normal.
   2. Inside the glass, the ray travels in a straight line at this new angle until it reaches the second, parallel surface.
   3. At the second surface, the ray crosses from glass back into air (less dense), so it speeds up again and bends away from the normal, by exactly the same angle it bent towards the normal on entry, because the two surfaces are parallel and the glass-to-air speed change is the exact reverse of the air-to-glass change.
   4. These two equal and opposite bends restore the ray's original direction, but because it travelled through the glass along a different path, it emerges laterally displaced from, though parallel to, its original line of travel. The answer is B.
   • Why not A: This wrongly assumes both refractions bend the ray in the same rotational sense. At the first surface the ray bends towards the normal on entering the denser glass, and at the second, parallel surface, where the ray is now leaving the denser glass into less dense air, it bends away from the normal by the same angle it originally bent towards it. These two equal but opposite bends cancel out, restoring the original direction, rather than doubling any single change.
   • Why not C: This confuses refraction with reflection; refraction changes the direction of a ray as it crosses into a medium where it travels at a different speed, but it does not reverse the ray's direction of travel by 180 degrees. Total reversal describes reflection straight back along the normal, not refraction through a transparent block.
   • Why not D: This correctly identifies that the ray emerges parallel to its original direction, but wrongly claims there is no overall shift. Because the ray travels through the glass at a different angle to its original path (having bent towards the normal on entry), it covers a different sideways distance while inside the block than it would have travelling in a straight line, so it emerges displaced sideways, though parallel to, its original path.

6. Question 6Answer: A

   1. The pulse travels from the ship down to the seabed and then back up to the ship, so the total distance it covers is twice the depth.
   2. Using distance = speed x time, the total distance travelled (down and back) is 1500 m/s x 0.8 s = 1200 m.
   3. Since this 1200 m covers the trip down and back, the depth itself is half of this: 1200 / 2 = 600 m.
   4. The answer is A.
   • Why not B: This finds the total distance travelled by the pulse (down to the seabed and back up again) as speed x time = 1500 x 0.8 = 1200 m, but forgets that this figure covers the round trip, giving it directly as the depth instead of halving it to isolate the one-way distance down to the seabed.
   • Why not C: This divides the total round-trip distance (1200 m) by 4 instead of by 2, as if the halving needed to be applied twice, when only a single halving is needed to convert the there-and-back distance into the one-way depth.
   • Why not D: This multiplies the total round-trip distance (1200 m) by 2 again, as though the fact that the pulse travels there and back required an extra factor of 2 beyond speed x time, rather than requiring that same total to be halved to find the depth.

7. Question 7Answer: D

   1. Loudness of a sound wave is determined by its amplitude: a larger amplitude means a louder sound.
   2. Pitch of a sound wave is determined by its frequency: a higher frequency means a higher pitch.
   3. Wave P has twice the amplitude of wave Q, so wave P is louder. Wave P has half the frequency of wave Q, so wave P has a lower pitch.
   4. Amplitude and frequency are independent properties, so a wave can be both louder and lower-pitched than another at the same time. The answer is D.
   • Why not A: This wrongly assumes amplitude and frequency are linked, as though a bigger wave must also vibrate faster; amplitude and frequency are independent properties of a wave. Wave P's larger amplitude does make it louder, but its lower frequency (half that of wave Q) means it has a lower pitch, not a higher one.
   • Why not B: This has loudness and pitch's dependencies swapped: loudness depends on amplitude (how much the particles are displaced), not frequency, and pitch depends on frequency, not amplitude. Since wave P has the larger amplitude, it is the louder wave, not equally loud, even though it correctly has the lower pitch due to its lower frequency.
   • Why not C: This assumes a lower-frequency wave must always be quieter, but loudness is determined by amplitude, not frequency; a low-frequency, large-amplitude wave, like wave P, can be louder than a higher-frequency, smaller-amplitude wave, since these two properties vary independently of each other.

8. Question 8Answer: C

   1. Every electromagnetic wave obeys wave speed = frequency x wavelength, so frequency = wave speed / wavelength, and here the wave speed is the same constant c for both waves, since both travel through a vacuum.
   2. For the radio wave: frequency = (3 x 10^8) / 300 = (3 x 10^8) / (3 x 10^2) = 10^6 Hz, since the mantissas 3 / 3 = 1 and the powers of ten subtract: 8 - 2 = 6.
   3. For the gamma ray: frequency = (3 x 10^8) / (3 x 10^-11) = 10^19 Hz, since 3 / 3 = 1 and the powers of ten subtract: 8 - (-11) = 19.
   4. Dividing the gamma ray's frequency by the radio wave's frequency: 10^19 / 10^6 = 10^(19-6) = 10^13. The answer is C.
   • Why not A: This inverts the ratio, computing the radio wave's frequency divided by the gamma ray's frequency (10^6 / 10^19 = 10^-13) instead of the gamma ray's frequency divided by the radio wave's, which the question asks for.
   • Why not B: This multiplies the two frequencies together instead of dividing one by the other: 10^6 x 10^19 = 10^25, treating the question's 'divided by' as though it asked for a product.
   • Why not D: This correctly finds the gamma ray's own frequency (10^19 Hz) but stops there, giving it directly as the answer instead of dividing it by the radio wave's frequency as the question asks.

9. Question 9Answer: B

   1. The atomic number gives the number of protons: 13.
   2. The number of neutrons is the mass (nucleon) number minus the atomic number: 27 - 13 = 14.
   3. A neutral atom of this element would have 13 electrons, equal to its 13 protons. A charge of 3+ means the ion has 3 fewer electrons than protons, since removing negatively charged electrons makes the overall charge more positive: 13 - 3 = 10 electrons.
   4. The ion has 13 protons, 14 neutrons and 10 electrons, so the answer is B.
   • Why not A: This correctly finds the proton and neutron counts but ignores the charge entirely, treating the ion as if it were a neutral atom with equal numbers of protons and electrons (13 each), when the 3+ charge means the ion has 3 fewer electrons than protons, not an equal number.
   • Why not C: This mistakes the charge number itself (3) for the total number of electrons in the ion, rather than using the charge to work out how many electrons must have been removed from the neutral atom; a 3+ charge means 3 fewer electrons than protons, so the ion has 13 - 3 = 10 electrons, not 3 in total.
   • Why not D: This adds the charge to the proton count instead of subtracting it from it, treating a 3+ charge as though it meant 3 extra electrons were gained (as for a negative ion) rather than 3 electrons lost (as for a positive ion), giving 13 + 3 = 16 electrons instead of 13 - 3 = 10.

10. Question 10Answer: D

    1. An alpha particle carries away a mass number of 4 and an atomic number of 2, so alpha decay reduces the nucleus's mass number by 4 and its atomic number by 2: starting from thorium-232 (atomic number 90), this gives mass number 232 - 4 = 228 and atomic number 90 - 2 = 88.
    2. A beta particle is an electron created when a neutron in the nucleus converts into a proton; this does not change the mass number (the total number of nucleons is unchanged) but increases the atomic number by 1, since there is now one more proton: mass number stays 228, and atomic number becomes 88 + 1 = 89.
    3. After both decays, the final nucleus has mass number 228 and atomic number 89.
    4. The answer is D.
    • Why not A: This correctly applies the alpha decay once (232 - 4 = 228, 90 - 2 = 88) but then wrongly treats the SECOND decay as another alpha decay too, subtracting a further 4 from the mass number and 2 from the atomic number (228 - 4 = 224, 88 - 2 = 86) instead of applying beta decay's effect, which leaves the mass number unchanged and increases the atomic number by 1.
    • Why not B: This correctly tracks the atomic number through both decays (90 - 2 + 1 = 89) but forgets that the alpha decay also reduces the mass number by 4, leaving the mass number at its original value of 232 instead of the correct 228.
    • Why not C: This correctly finds the mass number and atomic number after the alpha decay alone (228, 88), but then applies the beta decay's effect on the atomic number backwards, subtracting 1 (88 - 1 = 87) instead of adding 1 (88 + 1 = 89), as beta decay actually requires.

11. Question 11Answer: A

    1. A charged particle or ray ionises an atom by knocking an electron out of it, transferring some of its own energy to the atom in the process.
    2. Alpha particles are large, heavily charged and interact very strongly with nearby atoms, so they ionise many atoms in a very short distance, losing all their energy (and so being absorbed) within a few centimetres of air or a sheet of paper.
    3. Gamma radiation is uncharged electromagnetic radiation and interacts only weakly with matter, so it ionises comparatively few atoms per unit distance and can travel through thick lead before being significantly absorbed. Beta radiation sits between the two on both properties.
    4. So the type with the greatest ionising power (alpha) has the least penetrating power, and the type with the least ionising power (gamma) has the greatest penetrating power. The answer is A.
    • Why not B: This assumes that strong ionising power and strong penetrating power go together, but they are inversely related: a particle that ionises strongly transfers its energy to the material very quickly, over a short distance, which is precisely why it does not penetrate far. Alpha radiation is the most strongly ionising of the three, but this is exactly why it is stopped by something as thin as a sheet of paper, not why it penetrates furthest.
    • Why not C: This wrongly assumes that carrying more energy per particle, which is true of gamma radiation as a high-frequency electromagnetic wave, means it interacts most strongly with matter. In fact gamma radiation interacts only weakly with the atoms it passes, which is exactly why it is the least ionising of the three and can travel through thick lead before being significantly absorbed; the two properties move in opposite directions, not the same one.
    • Why not D: This wrongly claims the two properties are unrelated and gives an incorrect ranking; ionising power and penetrating power are inversely related for all three types, with alpha most ionising and least penetrating, gamma least ionising and most penetrating, and beta in between on both properties, not at either extreme.

12. Question 12Answer: C

    1. The number of half-lives that have passed is the total time divided by the half-life: 15 days / 3 days per half-life = 5 half-lives.
    2. Each half-life, the fraction of undecayed nuclei remaining is halved, so after n half-lives the fraction remaining is (1/2)^n.
    3. After 5 half-lives: (1/2)^5 = 1/32.
    4. The answer is C.
    • Why not A: This corresponds to using only 4 half-lives instead of the correct 5; 15 days divided by a 3-day half-life gives exactly 5 half-lives, not 4, so the fraction remaining should be halved one more time than this option allows: (1/2)^4 = 1/16 undercounts by one halving.
    • Why not B: This correctly finds 5 half-lives and correctly computes (1/2)^5 = 1/32 as a single factor, but then multiplies this result by the number of half-lives (5) again, as though the halving needed to be applied once more for each half-life on top of the exponent already accounting for them, giving 5 x 1/32 = 5/32 instead of simply 1/32.
    • Why not D: This treats the fraction remaining as though it fell in a straight line, one nth for n half-lives, rather than halving repeatedly; radioactive decay is exponential, not linear. Each half-life halves whatever remains, so after 5 half-lives the fraction is (1/2)^5 = 1/32, not 1/5.

13. Question 13Answer: D

    1. Background radiation is not part of the radioactive source and does not decay, so it must first be subtracted from the total measured count rate to find the count rate due to the source alone: 144 - 24 = 120 counts per minute.
    2. The number of half-lives that have passed in 15 minutes, given a 5-minute half-life, is 15 / 5 = 3 half-lives.
    3. The source's own count rate halves three times: 120 -> 60 -> 30 -> 15 counts per minute.
    4. The Geiger counter still detects the constant background radiation throughout, so the total count rate recorded is the source's remaining contribution plus the background: 15 + 24 = 39 counts per minute. The answer is D.
    • Why not A: This ignores that background radiation does not decay, and simply halves the total measured count rate three times (144 -> 72 -> 36 -> 18), treating the constant background as though it were part of the decaying source.
    • Why not B: This correctly subtracts the background at the start (144 - 24 = 120) and correctly halves the source's own contribution three times (120 -> 60 -> 30 -> 15), but then forgets that the Geiger counter still detects the constant background radiation throughout, reporting only the source's remaining contribution instead of the total count rate the counter would actually record.
    • Why not C: This forgets to subtract the background before halving, treating the whole 144 counts per minute as if it all decayed (144 / 8 = 18), but then adds the background of 24 back on top at the end anyway, effectively counting the background's contribution twice over.

14. Question 14Answer: B

    1. A gamma ray is electromagnetic radiation, a packet of energy with no mass and no electric charge.
    2. When an excited nucleus emits a gamma ray, it is releasing surplus energy left over from an earlier decay, not ejecting any protons or neutrons.
    3. Because no nucleons (protons or neutrons) leave the nucleus during gamma emission, the mass number (the total count of nucleons) and the atomic number (the count of protons) are both unchanged by this step.
    4. The answer is B.
    • Why not A: This wrongly attributes the alpha decay's effect (losing 2 protons and 2 neutrons, reducing mass number by 4 and atomic number by 2) to the separate gamma emission that follows it; the gamma ray itself carries away only energy, not any nucleons, so it causes no further change to the mass or atomic number beyond what the earlier alpha decay already caused.
    • Why not C: This wrongly borrows beta decay's effect, a neutron converting to a proton and increasing the atomic number by 1, and applies it to gamma emission; gamma emission does not involve any conversion of nucleons into other nucleons, only the release of a photon carrying excess nuclear energy, so it changes neither the mass number nor the atomic number.
    • Why not D: This invents a nucleon loss that does not occur in gamma decay; unlike alpha or beta decay, gamma emission releases only a photon, a packet of electromagnetic energy with no mass and no charge, so no nucleon of any kind leaves the nucleus and both the mass number and atomic number stay exactly the same as after the preceding alpha decay.

15. Question 15Answer: C

    1. Water waves travel more slowly in shallower water than in deeper water, in the same way that light travels more slowly in an optically denser medium than in a less dense one.
    2. The frequency of a wave is set by whatever is generating it (the source), and does not change as the wave crosses into a region where it travels at a different speed; this is true for both water waves and light.
    3. Since wave speed = frequency x wavelength, and the frequency stays fixed while the speed decreases on entering the shallower water, the wavelength must decrease too, so that the equation still balances.
    4. The shallower water, where the wave slows down, is therefore analogous to an optically denser medium, in which light also slows down; the wave's frequency is unchanged and its wavelength decreases in both cases. The answer is C.
    • Why not A: This reverses the direction of the speed change: water waves slow down, not speed up, when moving into shallower water, just as light slows down entering an optically denser medium, not a less dense one. Getting the speed direction backwards here also wrongly identifies the shallower water as analogous to a denser medium in which light speeds up, when a denser medium is precisely where light slows down.
    • Why not B: This correctly has the wave slow down, but incorrectly concludes the wavelength increases as a result; since wave speed = frequency x wavelength and the frequency stays fixed, a decrease in speed must be accompanied by a decrease in wavelength, not an increase, for the equation to still balance. It also wrongly calls the shallower water optically 'less dense', when a decrease in wave speed is precisely the signature of a denser medium in this analogy.
    • Why not D: This wrongly assumes frequency can change when a wave crosses into a region where it travels at a different speed; frequency is fixed by the source producing the wave and stays constant as the wave crosses the boundary, for both water waves and light. It is the wavelength that changes to compensate for the speed change, not the frequency, and this option reverses which quantity is actually constant.