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Question 1

(a) Fig. 1.1 is a diagram of parts of a tooth.
(i) State the letters from Fig. 1.1 that identify two parts that are dissolved by acid during dental decay.
(ii) State the letter from Fig. 1.1 that identifies part of the nervous system.
(iii) State the type of organism that causes dental decay.
(iv) State the names of two different types of human teeth.
(b) Table 1.1 shows some information about deficiency of some of the components in the diet.
Complete Table 1.1.
(c) One risk factor for coronary heart disease is an unhealthy diet.
State two other risk factors for coronary heart disease.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B7.2 — Digestive system (Parts (a)(i), (a)(ii), (a)(iv))
• Topic B10.1 — Diseases and immunity (Part (a)(iii))
• Topic B7.1 — Diet (Part (b))
• Topic B9.2 — Heart (Part (c))

▶️ Answer/Explanation

(a)(i) A and B

Enamel (A) and dentine (B) are the hard tissues on the outer part of the tooth.
Acid produced by bacteria dissolves these mineralised layers during decay.

(a)(ii) F

F points to the nerve inside the pulp cavity at the root of the tooth.
This is the part of the tooth connected to the nervous system.

(a)(iii) Bacteria

Bacteria in the mouth feed on sugars from food.
They produce acid as a waste product, which causes dental decay.

(a)(iv) Any two from: incisor, canine, pre-molar, molar

Humans have four types of teeth, each adapted for a different function.
Incisors and canines are used for cutting/tearing; molars and premolars are used for grinding.

(b)

Protein deficiency causes kwashiorkor / marasmus.
Vitamin C deficiency causes scurvy.
Fibre deficiency causes constipation.
Vitamin D / calcium deficiency causes rickets / weak bones.

(c) Any two from: stress, smoking, genetic predisposition, age, gender

Coronary heart disease has several risk factors besides diet.
These include lifestyle factors (smoking, stress) and non-modifiable factors (age, gender, genetics).

Question 2

A student investigates the reaction between calcium carbonate, \(\text{CaCO}_3\), and dilute hydrochloric acid, \(\text{HCl}\).
Calcium chloride, \(\text{CaCl}_2\), water and carbon dioxide are made.
(a) Construct the balanced symbol equation for this reaction.
(b) Describe the test for carbon dioxide. Include the observation for a positive result.
(c) Fig. 2.1 shows the apparatus used.
The student does the experiment at five different temperatures. Table 2.1 shows the results of the experiment.
(i) State the temperature when the reaction is fastest.
(ii) Describe the relationship between the temperature and the rate of the reaction.
(d) The student does the experiment again at \(21\,^\circ\text{C}\).
They use the same amounts of calcium carbonate and dilute hydrochloric acid.
This time they use hydrochloric acid that is more concentrated.
The reaction is faster than when using dilute hydrochloric acid.
Explain why.
Use ideas about collisions between particles in your answer.
(e) Fig. 2.2 shows the energy level diagrams for two different reactions, A and B.
Reaction A and reaction B are done under the same conditions.
Reaction A happens faster than reaction B.
Explain why.
Use information from Fig. 2.2 in your answer.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C3.1 — Formulas (Part (a))
• Topic C12.5 — Qualitative analysis (Part (b))
• Topic C6.2 — Rate of reaction (Part (c))
• Topic C6.2 — Rate of reaction (Part (d))
• Topic C5.1 — Exothermic and endothermic reactions (Part (e))

▶️ Answer/Explanation

(a) \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\)

All formulae must be correct.
The equation must be balanced with a coefficient of 2 in front of \(\text{HCl}\).

(b) Test: limewater; Result: turns milky/cloudy

Bubbling the gas through limewater (calcium hydroxide solution) is the standard test for \(\text{CO}_2\).
A positive result is shown by the limewater turning milky or cloudy.

(c)(i) 59 °C

The reaction is fastest at the highest temperature tested, since it has the shortest time to collect \(50\,\text{cm}^3\) of gas (13 s).

(c)(ii) As the temperature increases, the rate of reaction increases

This is shown by the decreasing time taken to collect the same volume of gas as temperature rises.

(d)

A more concentrated acid has more particles per \(\text{cm}^3\).
This means particles are more crowded, leading to more frequent collisions per second.
More frequent collisions between reacting particles increases the rate of reaction.

(e)

The activation energy for reaction A (\(15\,\text{kJ}\)) is lower than for reaction B (\(25\,\text{kJ}\)).
A lower activation energy means a higher proportion of reactant particles have enough energy to react on collision.
This makes reaction A proceed faster than reaction B under the same conditions.

Question 3

Fig. 3.1 shows an iceberg floating in the sea.
(a) The density of the iceberg is \(920\,\text{kg/m}^3\) and the volume of the iceberg is \(2 \times 10^5\,\text{m}^3\).
Calculate the mass of the iceberg.
(b)(i) Some samples of ice are taken from the iceberg so that a scientist can study what happens when the samples melt.
The scientist records the masses of three pieces of ice.
The pieces of ice are placed on top of blocks made of different materials.
The blocks are the same shape and size and are placed in a warm room so that they are all at the same temperature.
Fig. 3.2 shows the materials used.
After 5 minutes, the mass of each piece of solid ice remaining is measured. Table 3.1 shows the scientist’s results.
Use Fig. 3.2 and Table 3.1 to describe and explain the results of the scientist’s investigation.
(ii) Liquid water can be boiled to produce steam.
Describe the process of boiling in terms of the:
  • forces between molecules
  • distances between molecules
  • motion of molecules.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.4 — Density (Part (a))
• Topic P2.3.1 — Conduction (Part (b)(i))
• Topic P2.2.2 — Melting, boiling and evaporation (Part (b)(ii))

▶️ Answer/Explanation

(a) \(1.84 \times 10^8\,\text{kg}\)

Mass is calculated using \(m = \rho \times V\).
\(m = 920 \times (2 \times 10^5) = 1.84 \times 10^8\,\text{kg}\).

(b)(i)

The change in mass over 5 minutes is: polystyrene \(2.04\,\text{g}\), copper \(4.12\,\text{g}\), glass \(3.35\,\text{g}\).
Copper melts the ice the most, and polystyrene melts the ice the least.
This is because copper is a good conductor of heat, while polystyrene is a poor conductor (an insulator) containing trapped air, so it transfers the least energy to the ice.

(b)(ii)

During boiling, the forces of attraction between molecules decrease.
The distances between molecules increase as the liquid turns to gas.
The molecules become free to move and move out of the container, escaping as vapour.

Question 4

(a) Fig. 4.1 is a graph showing the effect of temperature on the rate of transpiration in one leaf.
(i) Complete the sentences to describe and explain the results shown in Fig. 4.1.
As the leaf temperature increases, the rate of transpiration increases.
Higher leaf temperatures result in increased ………………………… of water at the surfaces of the mesophyll cells.
This causes an increase in the rate water vapour ………………………… out of the leaf.
Water vapour is lost from the leaf through ………………………… in the lower epidermis.
(ii) The investigation is repeated at a greater humidity.
Draw a line on Fig. 4.1 to show the effect of greater humidity on the results.
(b) Xylem vessels transport water to the leaves.
(i) Explain the mechanism that causes the movement of water up the xylem.
(ii) State one other function of xylem apart from transport.
(c) State the names of two substances that are only transported in phloem.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B8.3 — Transpiration (Part (a))
• Topic B8.1 — Xylem and phloem (Part (b))
• Topic B8.4 — Translocation (Part (c))

▶️ Answer/Explanation

(a)(i) evaporation; diffuses; stomata

Higher temperature increases evaporation of water at the mesophyll cell surfaces.
This increases the rate at which water vapour diffuses out of the leaf.
Water vapour exits the leaf through the stomata in the lower epidermis.

(a)(ii)

A line should be drawn showing a decreased rate of transpiration compared to the original graph, since greater humidity reduces the diffusion gradient for water vapour leaving the leaf.

(b)(i)

Transpiration from the leaves creates a water potential gradient, reducing water potential at the top of the xylem.
This creates a “transpiration pull” that draws a column of water molecules upward.
Water molecules are held together by cohesion, allowing them to be pulled up as a continuous column.

(b)(ii) Support

Xylem tissue also provides mechanical support to the plant, helping it stay upright.

(c) Sucrose; amino acids

Phloem transports the products of photosynthesis (sucrose) and products of protein metabolism (amino acids), unlike xylem which only carries water and dissolved minerals.

Question 5

Lithium, sodium and potassium are metals in Group I of the Periodic Table.
(a) Describe the trend in reactivity of the Group I elements down the group.
(b) Table 5.1 shows some information about Group I elements.
Complete Table 5.1 by predicting the melting point of potassium and the density of rubidium.
Use ideas about trends down the group to help you.
(c) State the colour of the flame when sodium burns in oxygen. Tick (✓) one box.
(d) Potassium reacts with water.
Potassium hydroxide solution and hydrogen are made.
Complete the balanced equation for the reaction.
Include state symbols.
\(2\text{K(s)} + 2\text{H}_2\text{O(l)} \rightarrow …………. , ……. + ………… , …….
(e) Chlorine and bromine are elements in Group VII of the Periodic Table.
Chlorine displaces bromine from aqueous sodium bromide.
\(\text{Cl}_2 + 2\text{NaBr} \rightarrow \text{Br}_2 + 2\text{NaCl}\)
Explain why this is an example of a redox reaction.
(f) State which of the following is the electronic structure of an element in Group VIII (Group 0). Tick (✓) one box.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C8.2 — Group I properties (Part (a))
• Topic C8.2 — Group I properties (Part (b))
• Topic C12.5 — Qualitative analysis (Part (c))
• Topic C3.1 — Formulas (Part (d))
• Topic C8.3 — Group VII properties (Part (e))
• Topic C8.5 — Noble gases (Part (f))

▶️ Answer/Explanation

(a) Reactivity increases down the group

Going down Group I, the outer electron is further from the nucleus and more shielded.
This makes it easier to lose, so reactivity increases down the group.

(b) Melting point of potassium ≈ 64 °C (accepted range 38–97 °C); density of rubidium ≈ 1.63 g/cm³ (accepted range 0.90–1.92 g/cm³)

Melting point decreases going down Group I.
Density generally increases going down the group.
These values are predicted by following the trend between the surrounding elements.

(c) Yellow

Sodium compounds produce a characteristic yellow flame in flame tests, including when sodium burns in oxygen.

(d) \(2\text{K(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{KOH(aq)} + \text{H}_2\text{(g)}\)

The formulae must be correct and balanced.
State symbols (aq) for potassium hydroxide and (g) for hydrogen must be included.

(e)

A redox reaction involves the transfer of electrons — loss and gain occurring simultaneously.
The bromide ions (\(\text{Br}^-\)) lose electrons (oxidation) while chlorine atoms gain electrons (reduction).

(f) 2.8.8

Group VIII (Group 0) elements have a full outer shell.
Argon has the electronic structure 2.8.8, representing a complete octet.

Question 6

Fig. 6.1 shows wind turbines used to generate electricity.
(a) Fig. 6.2 shows how the power output of one wind turbine changes with wind speed.
On one particular day, the wind speed is \(10\,\text{m/s}\).
Calculate the energy generated by one wind turbine in 1 hour (3600 seconds).
(b) The wind turbine uses a generator to produce electricity.
Fig. 6.3 shows a simple a.c. generator.
(i) Describe how a simple a.c. generator produces a voltage output.
(ii) On Fig. 6.4, sketch a graph of voltage output against time for a simple a.c. generator rotating with a constant speed.
(c) Turbines and generators can also be used to convert the kinetic energy of tidal water into electrical energy.
(i) The efficiency of a tidal generator is 80% when the tidal water moves at \(5.0\,\text{m/s}\).
Calculate the mass of water which would need to pass through the tidal generator to produce \(1400\,\text{J}\) of electrical energy from kinetic energy.
(ii) State one advantage of using tidal generators to produce electricity instead of traditional fossil fuel power stations.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.6.4 — Power (Part (a))
• Topic P4.5.2 — The a.c. generator (Part (b))
• Topic P1.6.1 — Energy (Part (c)(i))
• Topic P1.6.3 — Energy resources (Part (c)(ii))

▶️ Answer/Explanation

(a) \(1.8 \times 10^8\,\text{J}\)

From Fig. 6.2, power output at \(10\,\text{m/s}\) is \(50\,\text{kW} = 50\,000\,\text{W}\).
Energy = power × time: \(E = 50\,000 \times 3600 = 1.8 \times 10^8\,\text{J}\).

(b)(i)

The coil turns/rotates within the magnetic field.
This changes the magnetic field passing through the coil (the coil cuts the field lines).
This induces an e.m.f./voltage output across the coil.

(b)(ii)

A sinusoidal wave should be sketched.
It must have constant amplitude and a constant time period, since the generator rotates at constant speed.

(c)(i) 140 kg

Kinetic energy needed: \(KE = 1400 / 0.8 = 1750\,\text{J}\).
Using \(KE = \tfrac{1}{2}mv^2\): \(m = \dfrac{2 \times 1750}{5.0^2} = \dfrac{3500}{25} = 140\,\text{kg}\).

(c)(ii)

Tidal generators do not release \(\text{CO}_2\)/greenhouse gases and do not contribute to global warming/climate change.
They also use a renewable energy source, unlike fossil fuels.

Question 7

(a) Albinism is an inherited condition that results in no pigments being made in the skin.
The allele for albinism is recessive a.
The allele for no albinism is dominant A.
Fig. 7.1 is a pedigree chart diagram of albinism in one family.
(i) Use the information in Fig. 7.1 to state:
the genotype of person 4
the genotype of person 1
the sex chromosomes of person 5.
(ii) A couple without albinism decides to have a child.
Complete the genetic diagram in Fig. 7.2 to calculate the percentage chance of having a child with albinism.
(b) Mitosis and meiosis are two forms of cell division.
(i) State two roles of mitosis.
(ii) State the name of one organ in the human body where meiosis occurs.
(iii) An organism has 32 chromosomes. State the number of chromosomes in a cell formed by meiosis in this organism.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B16.3 — Monohybrid inheritance (Part (a))
• Topic B16.2 — Cell division (Part (b))

▶️ Answer/Explanation

(a)(i) Person 4: aa; Person 1: Aa; Person 5: XX

Person 4 has albinism, so must be homozygous recessive (aa).
Person 1 does not have albinism but must carry the recessive allele since her son (person 4) is affected, giving genotype Aa.
Person 5 is female, so her sex chromosomes are XX.

(a)(ii) 25%

Both unaffected parents must be carriers (Aa) since albinism is recessive.
Crossing Aa × Aa gives offspring ratio 1 AA : 2 Aa : 1 aa.
So the percentage chance of a child having albinism (aa) is 25%.

(b)(i) Any two from: growth, repair of damaged tissues, replacement of cells, asexual reproduction

Mitosis produces genetically identical cells used for these processes in the body.

(b)(ii) Testes / ovary

Meiosis occurs in the reproductive organs to produce gametes.

(b)(iii) 16

Meiosis halves the chromosome number.
An organism with 32 chromosomes produces gametes with 16 chromosomes.

Question 8

(a) Petroleum is separated into different fractions.
Fig. 8.1 shows the percentage composition of fractions from a sample of petroleum.
(i) State one use of bitumen.
(ii) \(225\,\text{kg}\) of the sample of petroleum is placed into a barrel.
Calculate the mass of diesel oil, in kilograms, in this barrel.
(b) Petroleum is separated into different fractions by fractional distillation.
Describe how petroleum is separated by fractional distillation.
(c) Diesel oil, gasoline and other fuels made from petroleum naturally contain some sulfur impurities.
Suggest why sulfur impurities are removed from these fuels before the fuels are used.
(d) Gasoline used in cars causes air pollution by producing oxides of nitrogen such as nitrogen monoxide, NO.
Describe how a catalytic converter removes nitrogen monoxide from exhaust emissions.
Include a balanced symbol equation in your answer.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C11.3 — Fuels (Part (a))
• Topic C12.4 — Separation and purification (Part (b))
• Topic C10.2 — Air quality and climate (Part (c))
• Topic C10.2 — Air quality and climate (Part (d))

▶️ Answer/Explanation

(a)(i) Road surfaces

Bitumen, the thickest and least volatile fraction, is used for surfacing roads and roofing.

(a)(ii) 59 kg

Diesel oil makes up 26% of the petroleum sample.
Mass of diesel oil \(= \dfrac{225 \times 26}{100} = 58.5 \approx 59\,\text{kg}\).

(b)

Petroleum is heated, causing the different hydrocarbon fractions to vaporise and rise up the fractionating column.
Fractions are separated according to their boiling point, with fractions of lower boiling point rising higher.
As vapours rise and cool, they condense back to liquid at different heights depending on chain length.

(c)

Sulfur impurities are removed because burning them produces sulfur dioxide, which causes acid rain.

(d)

A catalytic converter contains a hot catalyst that helps nitrogen monoxide react and be removed from exhaust emissions.
This can occur via reaction with carbon monoxide or decomposition of NO itself.
Equation: \(2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2\) (or \(2\text{NO} \rightarrow \text{N}_2 + \text{O}_2\)).

Question 9

(a) Fig. 9.1 shows a butterfly resting on a leaf attached to the branch of a tree.
(i) State the name of the force labelled F.
(ii) The leaf will break off the branch if the moment about the pivot point X is greater than \(0.14\,\text{N cm}\).
The leaf does not break off the branch when the butterfly rests on it.
Calculate the maximum mass of the butterfly.
The gravitational field strength, \(g\), is \(10\,\text{N/kg}\).
(b) A scientist captures the butterfly in a plastic container to study it more closely.
The scientist places a converging lens across the top of the plastic container.
Fig. 9.2 shows the butterfly in the container.
Complete Fig. 9.3 to show how a thin converging lens forms a real image.
Label the image with the word image.
(c) The scientist uses a filament lamp to illuminate the butterfly while she is studying it.
(i) The filament lamp is in a series circuit with a cell and a switch.
Complete Fig. 9.4 to show this circuit.
(ii) Fig. 9.5 shows the current–voltage characteristic of a filament lamp.
Use Fig. 9.5 to explain how the resistance of the filament lamp changes as the voltage across it is increased.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.5.2 — Turning effect of forces (Part (a))
• Topic P3.2.3 — Thin converging lens (Part (b))
• Topic P4.3.1 — Circuit diagrams and circuit components (Part (c)(i))
• Topic P4.2.4 — Resistance (Part (c)(ii))

▶️ Answer/Explanation

(a)(i) Weight

The downward force F acting on the leaf due to the butterfly’s mass is its weight.

(a)(ii) 0.0028 kg

Weight = moment ÷ distance = \(0.14 / 5.0 = 0.028\,\text{N}\).
Mass = weight ÷ g = \(0.028 / 10 = 0.0028\,\text{kg}\).

(b)

A ray parallel to the axis is drawn refracting through the principal focus.
A second ray through the centre of the lens is drawn passing straight through undeviated.
Where the two rays cross (below the axis, inverted) marks the real image, which should be labelled image.

(c)(i)

The lamp, cell and switch should be drawn using correct standard circuit symbols.
All components must be connected in a single series loop.

(c)(ii)

As voltage increases, current increases.
Initially the line is straight (constant gradient), so resistance is constant.
The line then curves, showing the gradient reduces and resistance increases as the filament’s temperature rises.

Question 10

(a) The pH of the fluid in muscles changes during vigorous exercise due to the changing concentrations of lactic acid.
Table 10.1 shows the difference in pH before and immediately after vigorous exercise.
(i) Calculate the decrease in pH shown in Table 10.1.
(ii) Explain why there was a decrease in pH of the muscles during vigorous exercise.
(b) Muscle cells are adapted for movement as they are able to contract.
(i) Define the term movement.
(ii) State the name of the cell adapted for:
antibody production
movement of mucus
photosynthesis.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic B12.1 — Respiration (Part (a))
• Topic B1.1 — Characteristics of living organisms (Part (b)(i))
• Topic B2.1 — Cell structure (Part (b)(ii))

▶️ Answer/Explanation

(a)(i) 0.50

Decrease in pH \(= 7.08 – 6.58 = 0.50\).

(a)(ii)

During vigorous exercise there is not enough oxygen for aerobic respiration.
This causes the muscles to respire anaerobically.
Anaerobic respiration produces lactic acid, which reduces the pH of the muscle fluid.

(b)(i)

Movement is an action by an organism or part of an organism.
It causes a change of position or place.

(b)(ii) White blood cell; ciliated cell; palisade (mesophyll) cell

White blood cells (lymphocytes) produce antibodies.
Ciliated cells move mucus using their cilia.
Palisade mesophyll cells are adapted for photosynthesis, being packed with chloroplasts.

Question 11

(a) Element X is found in Group II of the Periodic Table.
State the formula of the ion formed by element X.
Tick (✓) one box.
(b) Determine the formula of the compound formed by \(\text{NH}_4^{+}\) and \(\text{CO}_3^{2-}\) ions.
(c) The number of subatomic particles in an ion is different from the number in a neutral atom.
Table 11.1 shows information about two different ions.
Complete Table 11.1.
(d) Carbon has the electronic structure 2.4.
Oxygen has the electronic structure 2.6.
Carbon reacts with oxygen to make carbon dioxide, \(\text{CO}_2\).
Complete the dot‑and‑cross diagram in Fig. 11.1 to show the bonding in carbon dioxide.
Only show the outer‑shell electrons.
(e) Carbon exists in several different forms.
Graphite and diamond are two of these forms.
Fig. 11.2 shows the structures of graphite and diamond.
(i) Explain why the structure of graphite makes it suitable for use as a lubricant.
(ii) Explain why the structure of diamond makes it suitable for use in cutting tools.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C2.4 — Ions and ionic bonds (Part (a))
• Topic C3.1 — Formulas (Part (b))
• Topic C2.2 — Atomic structure and the Periodic Table (Part (c))
• Topic C2.5 — Simple molecules and covalent bonds (Part (d))
• Topic C2.6 — Giant covalent structures (Part (e))

▶️ Answer/Explanation

(a) \(\text{X}^{2+}\)

Group II elements lose their two outer-shell electrons to form a 2+ ion.

(b) \((\text{NH}_4)_2\text{CO}_3\)

Two \(\text{NH}_4^{+}\) ions are needed to balance the 2− charge of one \(\text{CO}_3^{2-}\) ion, giving ammonium carbonate.

(c)

For \(\text{Al}^{3+}\): protons = 13, neutrons = mass number − protons = \(27 – 13 = 14\), electrons = 10 (given).
For \(\text{F}^{-}\): protons = 9, neutrons = 10 (given), electrons = protons + 1 = 10 (extra electron for negative charge).

(d)

Carbon forms a double covalent bond with each oxygen atom, sharing two pairs of electrons per bond.
Only the outer-shell (valence) electrons are shown, with dots and crosses representing electrons from each atom.

(e)(i)

Graphite has weak forces between its layers.
This allows the layers to slide over each other easily, making it useful as a lubricant.

(e)(ii)

Diamond has many strong covalent bonds arranged in a giant covalent (macromolecular/tetrahedral) structure.
This makes it extremely hard, ideal for cutting tools.

Question 12

X‑rays are part of the electromagnetic spectrum.
Hospitals use X‑rays for medical imaging.
(a)(i) State the speed of X‑rays.
(ii) An X‑ray machine in a hospital uses X‑rays with a wavelength of \(2.0 \times 10^{-11}\,\text{m}\).
Calculate the frequency of these X‑rays.
(b) Hospitals also use ultrasound waves for medical imaging.
(i) Ultrasound waves are high frequency sound waves which are longitudinal.
X‑rays are transverse waves.
Complete the sentences to describe the nature of longitudinal and transverse waves.
Longitudinal waves are produced by vibrations that are …………….. to the direction of energy transfer.
Transverse waves are produced by vibrations that are …………… to the direction of energy transfer.
(ii) During an ultrasound scan, ultrasound waves travel through gaseous air, solid bone and liquid blood.
Sound waves, including ultrasound waves, travel at different speeds in gases, solids and liquids.
Place the speed of sound in a gas, a solid and a liquid in order from fastest to slowest.
(c) Hospitals use radioactive tracers such as technetium‑99 (\(^{99}_{43}\text{Tc}\)) for medical imaging.
(i) \(^{99}_{43}\text{Tc}\) has a half‑life of 6 hours. Calculate the percentage of \(^{99}_{43}\text{Tc}\) remaining in a sample after 24 hours.
(ii) \(^{99}_{43}\text{Tc}\) is produced in hospitals from molybdenum‑99 (\(^{99}_{42}\text{Mo}\)). Use the correct nuclide notation to complete the decay equation for molybdenum‑99.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P3.3 — Electromagnetic spectrum (Part (a))
• Topic P3.1 — General properties of waves (Part (b)(i))
• Topic P3.4 — Sound (Part (b)(ii))
• Topic P5.2.4 — Half-life (Part (c)(i))
• Topic P5.2.3 — Radioactive decay (Part (c)(ii))

▶️ Answer/Explanation

(a)(i) \(3 \times 10^8\,\text{m/s}\)

X-rays, like all electromagnetic waves, travel at the speed of light in a vacuum.

(a)(ii) \(1.5 \times 10^{19}\,\text{Hz}\)

Using \(f = v/\lambda\): \(f = \dfrac{3 \times 10^8}{2.0 \times 10^{-11}} = 1.5 \times 10^{19}\,\text{Hz}\).

(b)(i) Longitudinal: parallel; Transverse: perpendicular

Longitudinal waves have particle vibrations parallel to the direction of energy transfer.
Transverse waves have particle vibrations perpendicular to the direction of energy transfer.

(b)(ii) Fastest: solid, liquid, slowest: gas

Sound travels fastest through solids (particles closely packed and strongly bonded), then liquids, and slowest through gases (particles far apart).

(c)(i) 6.25%

24 hours ÷ 6 hours per half-life = 4 half-lives.
Percentage remaining \(= \left(\dfrac{1}{2}\right)^4 \times 100\% = 6.25\%\).

(c)(ii) \(^{0}_{-1}\beta\)

Molybdenum-99 decays by beta emission: \(^{99}_{42}\text{Mo} \rightarrow \,^{99}_{43}\text{Tc} + \,^{0}_{-1}\beta\).
Mass number is conserved (99 = 99 + 0) and atomic number is conserved (42 = 43 + (−1)).

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