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CIE iGCSE Co-Ordinated Science P5.1 The nucleus Exam Style Questions Paper 3

Question

A house has an electric doorbell.
(a)(i) Draw a circuit diagram to show a doorbell connected in series with a switch and a battery.
Use the circuit symbol, , for the doorbell.
(ii) The battery has four 1.5 V cells in series. When the bell rings the current in the bell is 3.0 A.
Calculate the resistance of the bell.
(b) The house is fitted with a household fire (smoke) alarm.
The smoke detector in the alarm contains a radioactive isotope of americium-241 which emits α-particles.
(i) State the composition of an α-particle.
(ii) Americium-241 has a half-life of 430 years.
Suggest why the long half-life of americium-241 is important for use in a smoke detector.
(iii) Americium-241 has the nuclide notation \( ^{241}_{95}\text{Am} \).
State the number of neutrons in the nucleus of an atom of americium-241.

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

• Topic P4.3.1 — Circuit diagrams and circuit components (Part (a)(i))
• Topic P4.2.4 — Resistance (Part (a)(ii))
• Topic P5.2.2 — The three types of nuclear emission (Part (b)(i))
• Topic P5.2.4 — Half-life (Part (b)(ii))
• Topic P5.1 — The nucleus (Part (b)(iii))

▶️ Answer/Explanation

(a)(i) Correct symbol for battery and switch; series connections
The circuit diagram should show a battery, a switch, and the doorbell symbol connected in series with wires forming a complete loop.

(a)(ii) V = 4 × 1.5 OR 6.0 (seen); R = V/I (in any form) OR 6(.0)/3(.0); 2.0 (Ω)
Total voltage = 4 × 1.5 = 6.0 V. Using Ohm’s law, R = V/I = 6.0/3.0 = 2.0 Ω.

(b)(i) helium nucleus / 2 protons and 2 neutrons
An α-particle consists of 2 protons and 2 neutrons, which is identical to a helium-4 nucleus. It has a charge of +2e.

(b)(ii) decay rate needs to remain constant
A long half-life means the activity remains reasonably constant over the detector’s lifetime, so it does not need frequent replacement and remains reliable.

(b)(iii) 146
Number of neutrons = mass number – proton number = 241 – 95 = 146.

Question

Table 9.1 shows data about six metals.
Table 9.1 Data about six metals
(a)(i) Identify the metal in Table 9.1 that has the greatest density.
(a)(ii) Water has a density of $1000\,\text{kg/m}^3$.
Use data from Table 9.1 to explain why all the metals in Table 9.1 sink when placed in water.
(b)(i) Mercury is a liquid at room temperature (20 °C).
Explain how Table 9.1 shows this.
(b)(ii) Describe the structure of liquid mercury in terms of the arrangement and separation of the particles.
(b)(iii) Describe how the motion of particles in liquid mercury changes as the temperature decreases.
(c) Uranium-238 has the nuclide notation $^{238}_{92}\text{U}$.
Describe the composition of the nucleus of a uranium-238 atom.
(d) An alloy of lead and tin is used to make fuse wire.
The alloy has a melting point of 200 °C.
A fuse contains fuse wire and is used to protect electrical devices in electrical circuits.
Describe how the fuse protects the electrical circuit from the heating effect of an electric current.

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

• Topic P1.4 — Density (Part (a)(i) & (a)(ii))
• Topic P2.1.1 — States of matter (Part (b)(i))
• Topic P2.1.2 — Particle model (Part (b)(ii) & (b)(iii))
• Topic P5.1 — The nucleus (Part (c))
• Topic P4.4 — Electrical safety (Part (d))

▶️ Answer/Explanation

(a)(i)
uranium (density = $19\,100\,\text{kg/m}^3$)
Density is defined as mass per unit volume. Comparing all six metals in Table 9.1, uranium has the highest density value, making it the densest metal listed.

(a)(ii)
All the metals have a density greater than that of water ($1000\,\text{kg/m}^3$).
An object sinks in a fluid when its density is greater than the density of the fluid. The least dense metal in the table is aluminium at $2700\,\text{kg/m}^3$, which is still nearly three times the density of water. Since every metal listed exceeds $1000\,\text{kg/m}^3$, all will sink.

(b)(i)
Mercury’s melting point (−39 °C) is below 20 °C, and its boiling point (357 °C) is above 20 °C.
Room temperature (20 °C) lies between the melting point and the boiling point of mercury. This means mercury is above its melting point (not a solid) and below its boiling point (not a gas), so it must be a liquid at room temperature.

(b)(ii)
Arrangement: random / irregular — particles are not held in fixed positions.
Separation: close together, most touching.
Liquid particles have enough energy to overcome the rigid lattice of a solid, allowing them to move past one another, but they remain close together and are not widely spaced like gas particles. This gives liquids a fixed volume but no fixed shape.

(b)(iii)
The particles move more slowly as temperature decreases.
Temperature is a measure of the average kinetic energy of the particles. As thermal energy is removed, particles lose kinetic energy, their speed decreases, and collisions become less frequent and less energetic. If cooled sufficiently below its melting point of −39 °C, mercury would solidify.

(c)
92 protons and 146 neutrons
In the nuclide notation $^{238}_{92}\text{U}$, the lower number (92) is the proton number giving the number of protons, and the upper number (238) is the nucleon number (total protons + neutrons). The number of neutrons is therefore $238 – 92 = 146$.

(d)
1. If too large a current flows, the fuse wire heats up due to the heating effect of the current.
2. The temperature rises above 200 °C and the fuse wire melts.
3. This breaks the circuit, stopping all current flow and protecting the device.
The fuse is connected in series so that all current must pass through it. Its low-melting-point alloy ensures it melts before the current reaches a level that would overheat the wiring or damage the appliance, acting as a deliberate weak point that sacrifices itself to protect the rest of the circuit.

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