Home / 0625_m22_qp_42-Abigail

0625_m22_qp_42-Abigail

Question 1

A ball rolls down a ramp and onto a horizontal surface. The first section of the horizontal surface is smooth. The second section of the horizontal surface is rough. Fig. 1.1 shows a speed–time graph for the ball.
Fig 1.1 speed-time graph
(a) State the time when the ball reaches the start of the rough section of the horizontal surface.
(b) Explain how Fig. 1.1 shows that there is no resultant force on the ball when it rolls along the smooth section of the horizontal surface.
(c) Using Fig. 1.1, determine the acceleration of the ball as it rolls down the ramp.
(d) The ball starts from rest at the top of the ramp. Show that the length of the ramp is 9.6 m.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 1.2 — Motion: speed–time graphs, acceleration, distance from area under graph (Parts (a), (b), (c), (d))
• Topic 1.5.1 — Effects of forces: resultant force and Newton’s First Law (Part (b))

▶️ Answer/Explanation

(a)
time = 2.2 s
The rough section begins at the point on the graph where the ball transitions from constant speed to decelerating. Reading directly from the speed–time graph, this change in gradient occurs at t = 2.2 s, which is therefore when the ball first encounters the rough surface and a net friction force begins to act on it.

(b)
On the smooth section, the speed–time graph is a horizontal straight line, meaning the gradient is zero. Since acceleration equals the gradient of a speed–time graph, the acceleration is zero. By Newton’s Second Law (F = ma), a zero acceleration implies no resultant (net) force acts on the ball during this section of its journey.

(c)
acceleration = 8.5 m/s²
The acceleration on the ramp is found from the gradient of the straight-line section between t = 0 and t ≈ 1.5 s. Using a = Δv/Δt = 12.8/1.5 ≈ 8.5 m/s². A suitable triangle drawn on the ramp section of the graph with both coordinates clearly identified is required for full marks.

(d)
The length of the ramp equals the distance travelled during the ramp phase, which is the area under the speed–time graph between t = 0 and t = 1.5 s. This is a triangle: area = ½ × base × height = ½ × 1.5 × 12.8 = 9.6 m. Alternatively, using average velocity × time = 6.4 × 1.5 = 9.6 m, which confirms the given value.

Question 2

Fig. 2.1 shows a spring balance used to measure the weight of a baby. The spring inside the balance extends when a mass is suspended from it. The dial shows the extension of the spring as a value of mass in kg.
The spring obeys Hooke’s law up to a weight of 175 N.
(a)(i) State Hooke’s law. 
(a)(ii) State the relationship between the mass of the baby and the force exerted on the spring due to the baby. 
(a)(iii) The reading on the spring balance is 8.0 kg. Determine the force exerted on the spring due to the baby.
(b) The limit of proportionality for the spring is at a force of 175 N.
Sketch the extension–load graph for the spring. The sketch must continue beyond a force of 175 N. 
(c) The baby is carried from the ground floor to the bedroom. The vertical height of the bedroom above the ground floor is 3.5 m.
Calculate the change in gravitational potential energy of the baby when it is carried from the ground floor to the bedroom.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 1.5.1 — Effects of forces: Hooke’s law, load–extension graphs, limit of proportionality (Parts (a)(i), (a)(ii), (a)(iii), (b))
• Topic 1.3 — Mass and weight: W = mg (Part (a)(iii))
• Topic 1.7.1 — Energy: gravitational potential energy, ΔEp = mgΔh (Part (c))

▶️ Answer/Explanation

(a)(i)
Hooke’s law states that the extension of a spring is directly proportional to the applied force (load), provided the limit of proportionality is not exceeded. This means if the force doubles, the extension doubles — a relationship that holds only within the elastic region of the spring’s behaviour.

(a)(ii)
The force exerted on the spring is directly proportional to the mass of the baby, expressed by the equation W = mg. Since gravitational field strength g is constant (≈ 10 N/kg on Earth’s surface), doubling the mass of the baby doubles the downward force on the spring.

(a)(iii)
force = 80 N
Using W = mg, with m = 8.0 kg and g = 10 N/kg: W = 8.0 × 10 = 80 N. This is the gravitational force (weight) of the baby acting downward on the spring, causing it to extend by a corresponding amount that the dial reads as 8.0 kg.

(b)
The sketch should show a straight line through the origin with a positive gradient up to the limit of proportionality at 175 N, representing the region where Hooke’s law holds. Beyond 175 N, the line curves upward with an increasing gradient (the spring stretches more per unit force), indicating the spring no longer returns to its original length.

(c)
change in GPE = 280 J
Using ΔEp = mgΔh = 80 N × 3.5 m = 280 J. The weight of the baby (80 N) acts as the force, and the vertical height gained (3.5 m) is the displacement in the direction of that force. This energy is stored as gravitational potential energy in the baby–Earth system.

Question 3

Fig. 3.1 and Fig. 3.2 show how a puddle of water changes on a warm windy day. 

(a) Describe the process by which the volume of water in the puddle decreases.
(b) State and explain one change in the weather that would cause the volume of water in the puddle to decrease more slowly.
(c) Explain, in terms of molecules, how sweating helps to cool your body on a hot day.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 2.2.3 — Melting, boiling and evaporation: evaporation in terms of escape of energetic molecules, cooling effect of evaporation (Parts (a), (b), (c))
• Topic 2.2.3 — Supplement: effect of temperature, surface area and air movement on evaporation rate (Part (b))

▶️ Answer/Explanation

(a)
The volume of water decreases through the process of evaporation. At the surface of the puddle, the most energetic water molecules have sufficient kinetic energy to overcome intermolecular forces and escape into the surrounding air as water vapour. This process occurs continuously at temperatures below boiling point, and is accelerated on a warm, windy day because higher temperature increases average molecular energy and wind removes vapour from above the surface, allowing more molecules to escape.

(b)

Explanation: At a lower temperature, water molecules have less kinetic energy on average, so fewer molecules at the surface have enough energy to escape into the air as vapour — this reduces the rate of evaporation. Similarly, less wind means evaporated molecules accumulate above the puddle, reducing the concentration gradient and slowing further evaporation.

(c)
When you sweat, liquid sweat is deposited on the skin’s surface. Thermal energy from the body is transferred to the sweat molecules, increasing their kinetic energy. The most energetic sweat molecules escape from the surface of the skin into the air, leaving behind molecules with a lower average kinetic energy. Since temperature is proportional to average kinetic energy, the remaining sweat — and the skin in contact with it — cools down, effectively removing thermal energy from the body.

Question 4

(a) A sample of sand has a volume of 0.050 m³. The density of the sand is 1900 kg/m³. The specific heat capacity of the sand is 1500 J/(kg °C).
(a)(i) Calculate the mass of the sample of sand.
(a)(ii) Calculate the thermal capacity of the sample of sand.
(a)(iii) The initial temperature of the sample of sand is 7.0 °C. The sample of sand is heated using an electrical heater. The power of the heating element is 50 W.
Calculate the time taken to increase the temperature of the sand to 19.0 °C.
(b) In some countries, the soil is too cold for plants to grow well. Plants are grown in plastic pots placed on sand. The sand is heated using an electrical heater, as shown in Fig. 4.1.
(b)(i) Describe, in terms of molecules, how thermal energy is transferred from the heated sand through the base of the plastic pot.
(b)(ii) The heating element in Fig. 4.1 remains switched on. The temperature of the sand remains constant at a value above room temperature.
Explain why the temperature of the sand remains constant.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 1.4 — Density: ρ = m/V (Part (a)(i))
• Topic 2.2.2 — Specific heat capacity: c = ΔE/(mΔθ), thermal capacity (Parts (a)(ii), (a)(iii))
• Topic 2.3.1 — Conduction: thermal conduction in terms of molecular vibrations (Part (b)(i))
• Topic 2.3.3 — Radiation / Supplement: energy balance at constant temperature (Part (b)(ii))

▶️ Answer/Explanation

(a)(i)
mass = 95 kg
Using the density equation ρ = m/V, rearranged to m = ρV: m = 1900 × 0.050 = 95 kg. Density is defined as mass per unit volume, so multiplying the density of sand by its volume directly gives the mass of the sample. This value is then used in parts (ii) and (iii).

(a)(ii)
thermal capacity = 142 500 J/°C ≈ 1.4 × 10⁵ J/°C
Thermal capacity is defined as the energy required to raise the temperature of an object by 1 °C, calculated as thermal capacity = mass × specific heat capacity = 95 × 1500 = 142 500 J/°C. Unlike specific heat capacity, thermal capacity applies to the whole object rather than per unit mass, making it useful when working with a fixed sample.

(a)(iii)
time = 34 000 s
The temperature rise required is Δθ = 19.0 − 7.0 = 12.0 °C. The energy needed is E = thermal capacity × Δθ = 142 500 × 12.0 = 1 710 000 J. Using P = E/t, the time is t = E/P = 1 710 000 / 50 = 34 200 s ≈ 34 000 s (approximately 9 hours 27 minutes), assuming all electrical energy is transferred to the sand with no losses.

(b)(i)
When the sand is heated, its molecules gain kinetic energy and vibrate more vigorously. These molecules collide with neighbouring molecules in the base of the plastic pot, transferring energy through successive collisions from molecule to molecule. This process of conduction continues through the solid plastic, progressively transferring thermal energy from the hotter sand to the cooler soil in the pot above.

(b)(ii)
Although the heater continuously supplies energy to the sand, the sand is at a higher temperature than its surroundings and therefore continuously loses thermal energy to the environment by conduction, convection and radiation. The temperature remains constant because the rate of energy supplied by the heater equals the rate of energy lost to the surroundings — a state of thermal equilibrium is reached where energy input and output are balanced.

Question 5

A boy looks at the image of a clock in a plane mirror. Fig. 5.1 shows the mirror, the clock and the position of one of the boy’s eyes.
(a)(i) On Fig. 5.1, draw a ray of light from the clock, reflected to the boy’s eye. 
(a)(ii) On Fig. 5.1, mark with an X the position of the image of the clock. 
(a)(iii) State whether the image formed by the mirror is virtual or real. Explain your answer. 
(a)(iv) Fig. 5.2 shows the image of the clock seen by the boy in the mirror.
The boy now looks directly at the clock.
On Fig. 5.3, draw what the boy sees.
(b)(i) The clock is illuminated by a source of monochromatic green light.
State the meaning of monochromatic.
(b)(ii) The green light has a wavelength of 5.6 × 10⁻⁷ m.
Calculate the frequency of this green light.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 3.2.1 — Reflection of light: normal, angle of incidence, angle of reflection, plane mirror image characteristics (Parts (a)(i), (a)(ii), (a)(iii), (a)(iv))
• Topic 3.2.4 — Dispersion of light / Supplement: monochromatic light (Part (b)(i))
• Topic 3.1 — General properties of waves: wave speed equation v = fλ, speed of electromagnetic waves (Part (b)(ii))
• Topic 3.3 — Electromagnetic spectrum: speed of EM waves in vacuum = 3.0 × 10⁸ m/s (Part (b)(ii))

▶️ Answer/Explanation

(a)(i)
A correct ray diagram requires a straight line drawn from the clock to the mirror surface, and a second straight line from that point of incidence to the boy’s eye, with arrows showing the direction of travel of light. The law of reflection must be obeyed at the mirror surface — the angle of incidence must equal the angle of reflection, both measured from the normal to the mirror at the point of incidence.

(a)(ii)
The image of the clock in a plane mirror is located directly behind the mirror, at the same perpendicular distance behind the mirror as the clock is in front of it. The position X should be marked on the line extending perpendicularly from the clock through the mirror, at an equal distance on the other side — this is found by extending the reflected rays back behind the mirror until they meet.

(a)(iii)
The image is virtual. A virtual image is one that cannot be projected onto a screen because the light rays do not actually pass through the image position — they only appear to diverge from it when extended backwards behind the mirror. Since no real light rays converge at the image location, the image cannot be captured on a screen placed behind the mirror.

(a)(iv)


When looking directly at the clock (not in the mirror), the boy sees the clock the correct (non-reversed) way round. The image seen in the mirror appears laterally inverted — left and right are swapped — so the direct view of the clock shows the numbers and hands in their true, unreversed orientation. The drawing in Fig. 5.3 should therefore show a normal clock face, the mirror image of what was shown in Fig. 5.2.

(b)(i)
Monochromatic means the light consists of a single frequency (and therefore a single wavelength). Unlike white light, which is a mixture of many frequencies across the visible spectrum, monochromatic green light has only one specific frequency, which is why it appears as a pure, single colour with no dispersion when passed through a prism.

(b)(ii)
frequency = 5.4 × 10¹⁴ Hz
Using the wave equation v = fλ, rearranged to f = v/λ: f = (3.0 × 10⁸) / (5.6 × 10⁻⁷) = 5.357 × 10¹⁴ ≈ 5.4 × 10¹⁴ Hz. The speed of all electromagnetic waves in a vacuum (and approximately in air) is 3.0 × 10⁸ m/s, which must be recalled and substituted correctly alongside the given wavelength.

Question 6

Fig. 6.1 shows two bar magnets.
(a) On Fig. 6.1, sketch the pattern and the direction of the magnetic field lines between the bar magnets.
(b) Fig. 6.2 shows the same bar magnets with a coil of wire placed between them.
 
(b)(i) Name the parts labelled A in Fig. 6.2.
(b)(ii) The coil of wire is rotated in the direction shown in Fig. 6.2.
On Fig. 6.2, draw an arrow to show the direction of the current in the coil. Explain your answer.
(b)(iii) Explain how rotating the coil in Fig. 6.2 continuously causes the galvanometer needle to show an alternating current.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 4.1 — Simple phenomena of magnetism: magnetic field pattern and direction around bar magnets (Part (a))
• Topic 4.5.1 — Electromagnetic induction: conductor moving in a magnetic field induces e.m.f. (Part (b)(ii))
• Topic 4.5.2 — The a.c. generator: slip rings, brushes, rotating coil, alternating current (Parts (b)(i), (b)(ii), (b)(iii))

▶️ Answer/Explanation

(a)
The magnetic field lines between the two bar magnets should be drawn as curved lines running from the North pole to the South pole, with arrows indicating this direction. The central field line runs perpendicular and straight between the two poles. At least two additional field lines curving outward on either side should be shown, with closer spacing near the poles indicating a stronger field. Lines must not cross each other at any point.

(b)(i)
The parts labelled A are slip rings. Slip rings are continuous conducting rings attached to the ends of the rotating coil, and they maintain a permanent electrical connection between the rotating coil and the external circuit (including the galvanometer) via stationary carbon brushes that press against them. Unlike a split-ring commutator used in a d.c. motor, slip rings allow the direction of current to reverse, enabling alternating current to be delivered to the external circuit.

(b)(ii)


The direction of the induced current in the coil can be determined using Fleming’s Right-Hand Rule (for generators): the thumb points in the direction of motion of the conductor, the index finger points in the direction of the magnetic field, and the middle finger points in the direction of the induced current. The arrow on the coil should be drawn in the direction consistent with this rule for the given rotation direction, showing the current flowing in the correct sense around the coil at that instant.

(b)(iii)
As the coil rotates continuously in the magnetic field, its wire sides cut through the magnetic field lines, inducing an e.m.f. and hence a current in the coil. During the first half-rotation, one side of the coil moves upward through the field and the other moves downward, producing current in one direction. During the next half-rotation, the motion of each side reverses relative to the field direction, so the induced current reverses direction. This reversal occurs every half-turn, producing a continuously alternating current that causes the galvanometer needle to swing back and forth. The current is at its maximum when the coil sides move perpendicular to the field, and zero when the coil is parallel to the field (no field lines being cut).

Question 7

Fig. 7.1 shows a circuit including a 12 V battery, two identical lamps.
(a) The 12 V battery consists of cells connected in series. Each cell in the battery has an electromotive force (e.m.f.) of 1.5 V.
Determine how many cells are in the battery
(b)(i) When the switch is closed, the ammeter reading is 2.4 A.
Calculate the total resistance of the circuit.
(b)(ii) Each lamp has a resistance of 3.0 Ω.
Calculate the resistance of Q.
(c)(i) On Fig. 7.1, draw the symbol for a voltmeter that measures the potential difference (p.d.) across the two lamps. 
(c)(ii) Calculate the power supplied to one lamp.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 4.2.3 — Electromotive force and potential difference: e.m.f. of cells in series (Part (a))
• Topic 4.2.4 — Resistance: R = V/I (Part (b)(i))
• Topic 4.3.2 — Series and parallel circuits: combined resistance of parallel resistors, current and p.d. in series and parallel circuits (Parts (b)(ii), (c)(i))
• Topic 4.2.5 — Electrical energy and electrical power: P = IV (Part (c)(ii))

▶️ Answer/Explanation

(a)
number of cells = 8
When cells are connected in series, their individual e.m.f. values add together to give the total battery e.m.f. Dividing the total battery voltage by the e.m.f. of one cell: number of cells = 12 V ÷ 1.5 V = 8 cells. This is a straightforward application of the rule that e.m.f.s in series are cumulative, similar to how resistances in series add together.

(b)(i)
total resistance = 5.0 Ω
Using Ohm’s law in the form R = V/I, where V is the total e.m.f. of the battery and I is the total current measured by the ammeter: R = 12 / 2.4 = 5.0 Ω. This is the total resistance of the entire circuit, including both the resistor Q and the parallel combination of the two lamps in series with it.

(b)(ii)
resistance of Q = 3.5 Ω
The two identical lamps (each 3.0 Ω) are connected in parallel, so their combined resistance is found using 1/R = 1/3.0 + 1/3.0 = 2/3.0, giving R = 1.5 Ω. Since the total circuit resistance is 5.0 Ω and this is the sum of Q and the parallel lamp combination in series: resistance of Q = 5.0 − 1.5 = 3.5 Ω.

(c)(i)
A voltmeter is always connected in parallel with the component whose p.d. is being measured. On Fig. 7.1, the voltmeter symbol (a circle containing the letter V) should be drawn connected across both lamps — that is, with one terminal connected to the junction before the lamps and the other terminal connected to the junction after the lamps, forming a parallel branch alongside the lamp combination.

(c)(ii)
power = 4.3 W
The total current through the circuit is 2.4 A, which splits equally between the two identical lamps in parallel, giving 1.2 A through each lamp. The p.d. across the parallel lamp combination is V = IR = 1.2 × 3.0 = 3.6 V (or equivalently, total p.d. minus p.d. across Q: 12 − (2.4 × 3.5) = 3.6 V). Power for one lamp: P = IV = 1.2 × 3.6 = 4.32 W ≈ 4.3 W.

Question 8

A radio is connected to the mains supply using a step-down transformer.
(a) Draw a labelled diagram of the structure of a basic step-down transformer.
(b) Explain the operation of a basic transformer.
(c) The voltage of the mains supply is 230 V. The output voltage of the transformer is 6.0 V. Calculate the value of the turns ratio \(\left(\dfrac{N_s}{N_p}\right)\). Give your answer to two significant figures.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 4.5.6 — The transformer: structure of a simple transformer, step-up and step-down, turns ratio equation Vp/Vs = Np/Ns (Parts (a), (b), (c))
• Topic 4.5.6 Supplement — Principle of operation of a simple iron-cored transformer: alternating current, changing magnetic field, induced e.m.f. (Part (b))

▶️ Answer/Explanation

(a)
The diagram must show a soft iron (laminated) core forming a closed rectangular loop, with two separate coils wound around opposite sides of the core. The coil connected to the input (mains) supply should be labelled the primary coil, and the coil connected to the output (radio) should be labelled the secondary coil. Since this is a step-down transformer, the secondary coil must clearly show fewer turns than the primary coil. Both coils must be connected to separate external circuits and must not be electrically connected to each other — energy is transferred only through the shared magnetic field in the iron core.

(b)
An alternating current in the primary coil produces a continuously changing magnetic field in the soft iron core. This changing magnetic field links with the secondary coil, and by the principle of electromagnetic induction, a changing magnetic flux induces an e.m.f. (and hence an alternating current) in the secondary coil. The ratio of the induced secondary voltage to the primary voltage is equal to the ratio of the number of turns on the secondary coil to the number of turns on the primary coil, as described by the transformer equation. A transformer only works with alternating current because a steady (direct) current would produce a constant magnetic field that would not induce any e.m.f. in the secondary coil.

(c)
turns ratio Ns/Np = 0.026
Using the transformer equation: Vp/Vs = Np/Ns, which can be rearranged to give Ns/Np = Vs/Vp = 6.0 / 230 = 0.02608… ≈ 0.026 (to 2 significant figures). This value being less than 1 confirms that the secondary coil has fewer turns than the primary, consistent with a step-down transformer that reduces the mains voltage of 230 V down to the 6.0 V required by the radio.

Question 9

Fig. 9.1 shows a digital circuit.
(a)(i) Explain what is meant by digital.
(a)(ii) Table 9.1 is a truth table for the digital circuit shown in Fig. 9.1.
Complete the columns C and D in Table 9.1.
 
(b) State the single logic gate that would produce the same output D from inputs A and B.
▶️ Answer/Explanation

(a)(i)
A digital signal is one that can only take one of two possible states — represented as 0 (low/off) or 1 (high/on). Unlike analogue signals, which can take any value within a continuous range, digital signals are discrete and binary in nature. This makes them more resistant to noise and interference, which is why digital systems are widely used in modern electronics and communications.

(a)(ii)
Column C is the output of the NOT gate applied to input B, so C = NOT B. Column D is the output of the AND gate with inputs A and C, so D = A AND C. The completed truth table is:

The NOT gate inverts B before it enters the AND gate, so output D is only 1 when A is 1 and B is 0 — meaning the circuit outputs 1 only when A is HIGH and B is LOW simultaneously.

(b)
The single logic gate that produces the same truth table output D from inputs A and B directly is a NAND gate. Comparing the output column D with a standard NAND truth table for inputs A and B confirms this equivalence — the combination of a NOT gate on one input followed by an AND gate is logically equivalent to a single NAND gate, a result that can be verified by inspection of the completed truth table above.

Question 10

The isotope americium-241 is represented by:

\(_{95}^{241}\textrm{Am}\)

This isotope decays by α-emission to an isotope of neptunium (Np).

(a) Complete the nuclide equation for this decay. [3]

\(_{95}^{241}\textrm{Am} \rightarrow \, _{\ }^{\ }\textrm{Np} \; + \; _{\ }^{\ }\alpha\)

(b) Fig. 10.1 shows a simple diagram of a smoke detector. The smoke detector contains a small sample of americium-241. This isotope ionises the air between the metal plates in the detector, allowing a small current to flow. When smoke enters the detector, this current is interrupted and the alarm sounds.
 
(b)(i) Describe how the americium-241 ionises the air between the metal plates.
(b)(ii) Suggest and explain two reasons why smoke detectors use an isotope that emits α-particles rather than an isotope that emits γ-radiation.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic 5.1.2 — The nucleus: nuclide notation, proton number, nucleon number, decay equations (Part (a))
• Topic 5.2.2 — The three types of nuclear emission: nature, ionising effect and penetrating ability of α, β and γ radiation (Parts (b)(i), (b)(ii))
• Topic 5.2.4 — Half-life Supplement: how type of radiation and half-life determine suitability for applications including smoke alarms (Part (b)(ii))
• Topic 5.2.5 — Safety precautions: effects of ionising radiation on living things (Part (b)(ii))

▶️ Answer/Explanation

(a)
\(_{95}^{241}\textrm{Am} \rightarrow \, _{93}^{237}\textrm{Np} \; + \; _{2}^{4}\alpha\)
In alpha decay, the nucleus emits a helium-4 nucleus (2 protons and 2 neutrons). To conserve nucleon number: 241 = 237 + 4 ✓. To conserve proton number: 95 = 93 + 2 ✓. The daughter nucleus therefore has nucleon number 237 and proton number 93, which corresponds to neptunium (Np). Both conservation laws — conservation of nucleon number and conservation of proton (charge) number — must be satisfied simultaneously in any valid nuclear decay equation.

(b)(i)
The americium-241 emits alpha particles (helium-4 nuclei) that travel through the air between the metal plates. As these highly energetic, doubly-charged alpha particles move through the air, they interact strongly with air molecules, colliding with and knocking electrons out of the air molecules, thereby converting neutral air molecules into positive ions and free electrons. This creates pairs of oppositely charged ions in the air, allowing a small electric current to flow between the oppositely charged metal plates — a current that is interrupted when smoke particles absorb the alpha particles before they can ionise the air.

(b)(ii)
Reason 1: Alpha particles have a very short range in air (only a few centimetres) and are completely stopped by the plastic casing of the smoke detector. This means the radiation poses minimal danger to people living in the home, as it cannot penetrate through the detector’s outer casing and reach the human body. Gamma radiation, by contrast, is highly penetrating and would pass through the casing and through the human body, posing a significant ongoing radiation hazard to occupants.

Reason 2: Alpha particles are far more strongly ionising than gamma radiation, producing many more ion pairs per unit length of path through air. This makes alpha emitters much more effective at ionising the air between the plates to produce the small detection current needed for the alarm to function correctly. A gamma-emitting source of equivalent activity would ionise the air far too weakly to maintain a reliable detection current, making the smoke detector ineffective.

Scroll to Top