Secondary 3 Physics Modern Physics Quiz

Secondary 3 Physics Quiz – Modern Physics: ANSWER KEY

Total Marks: 40
Duration: 50 minutes

Section A: Multiple Choice (Questions 1–10)

Section A Total: 20 marks

Section B: Structured Response

11. [3 marks]

[1 mark per correct row; accept equivalent valid differences such as products, energy release per nucleon, or fuel types]

Common mistakes: Confusing which process happens where; stating that fusion occurs in reactors currently (commercial fusion is still experimental).

12. [4 marks]

(a) [2 marks]

Number of half-lives in 18 hours = 18 ÷ 6 = 3 half-lives [1]

Count rate after 3 half-lives = 4800 ÷ $2^3$ = 4800 ÷ 8 = 600 counts per minute [1]

Working must show: identification of 3 half-lives and correct calculation. Accept 4800 → 2400 → 1200 → 600 as alternative working.

(b) [1 mark]

3 half-lives

(c) [1 mark]

Radioactive decay is a random process [½]. Even after many half-lives, there is always a small probability that some nuclei have not yet decayed [½]. The activity approaches zero asymptotically but theoretically never reaches exactly zero.

13. [4 marks]

(a) [2 marks]

For conservation of mass number: 241 = $A$ + 4, so $A$ = 237 [1]

For conservation of charge (atomic number): 95 = $Z$ + 2, so $Z$ = 93 [1]

Completed equation: $^{241}_{95}\text{Am} \rightarrow ^{237}_{93}\text{Np} + \ ^4_2\alpha$

(b) [2 marks]

Any two of the following:

• Alpha particles have very low penetrating power [½] and are stopped by a few centimetres of air or the plastic casing of the detector [½]
• The americium is present in very small quantities sealed inside the smoke detector [½]
• The source is not removed from the detector during normal use [½]
• The distance from the source to anyone nearby is sufficient that alpha particles cannot reach them [½]

Maximum 2 marks.

14. [3 marks]

Working from the diagram (nuclear power station flow):

1. Nuclear energy → Thermal energy: Nuclear fission in fuel rods releases energy as heat in the reactor core [1]
2. Thermal energy → Kinetic energy: Heat boils water to produce steam, which drives the turbine [1]
3. Kinetic energy → Electrical energy: The turbine turns the generator, producing electricity [1]

For full marks, must mention all three transformations in correct order with correct energy forms. Accept "heat" for thermal energy and "mechanical energy" for kinetic energy.

15. [5 marks]

(a) [3 marks]

Number of half-lives = 4800 ÷ 1600 = 3 half-lives [1]

Atoms remaining = $1.6 \times 10^{24}$ ÷ $2^3$ [1]

= $1.6 \times 10^{24}$ ÷ 8 = $2.0 \times 10^{23}$ atoms [1]

Alternative working: $1.6 \times 10^{24} \rightarrow 8.0 \times 10^{23} \rightarrow 4.0 \times 10^{23} \rightarrow 2.0 \times 10^{23}$

(b) [2 marks]

• Even though the half-life is long, radium-226 is highly radioactive and emits alpha particles that are very ionizing [1]
• Long half-life means the source remains dangerous for thousands of years, posing long-term storage and contamination risks [1]
• Alpha emitters are particularly hazardous if ingested or inhaled, as the radiation is concentrated in tissue [1]

Maximum 2 marks.

Section C: Application and Analysis

16. [4 marks]

(a) [2 marks]

Reading from graph: Initial activity = 3200 counts/min [½]

After one half-life, activity falls to 1600 counts/min [½]

From graph, this occurs at 10 hours [1]

Accept 9–11 hours if clearly reading from drawn graph. Alternative: could use any two corresponding points, e.g., 1600 to 800 gives same half-life.

(b) [2 marks]

From 40 hours to 50 hours is one half-life (since half-life = 10 hours) [1]

Activity at 40 hours = 200 counts/min

Activity after 50 hours = 200 ÷ 2 = 100 counts per minute [1]

Alternative: 50 hours = 5 half-lives, so activity = 3200 ÷ $2^5$ = 3200 ÷ 32 = 100 counts/min

17. [4 marks]

(a) [2 marks]

Ratio remaining = $\frac{1}{4}$ of original [½]

Number of half-lives needed: $\left(\frac{1}{2}\right)^n = \frac{1}{4}$, so $n = 2$ [1]

Age of artifact = 2 × 5700 = 11 400 years [½]

(b) [2 marks]

• After 65 million years, the number of half-lives = 65 000 000 ÷ 5700 ≈ 11 400 half-lives [1]
• This means the carbon-14 would have decayed to essentially undetectable levels [½]
• The remaining carbon-14 would be indistinguishable from background radiation or contamination [½]
• For such old samples, isotopes with longer half-lives (e.g., potassium-40, uranium-238) are used instead [½]

Maximum 2 marks for part (b).

18. [3 marks]

Any three valid precautions:

Maximum 3 marks. Must include explanation linked to physics for full mark each.

19. [4 marks]

Detection arrangement: [1 mark for any valid setup]

A Geiger-Müller tube is placed at a fixed distance from the unknown source. Counts are recorded with:

• No absorber present
• Paper inserted between source and detector
• 5 mm aluminium inserted
• 50 mm lead inserted

Identification method: [3 marks]

If counts remain unchanged through all absorbers: source emits gamma (and very energetic gamma if not reduced by lead). If no counts at all: source may be too weak, too distant, or not radioactive.

20. [4 marks]

(a) [2 marks]

Mass of 4 hydrogen nuclei = 4 × $1.673 \times 10^{-27}$ kg = $6.692 \times 10^{-27}$ kg [½]

Mass of products = $6.647 \times 10^{-27}$ + (2 × $9.11 \times 10^{-31}$) [½]

= $6.647 \times 10^{-27}$ + $1.822 \times 10^{-30}$ = $6.648822 \times 10^{-27}$ kg [½]

Mass defect = $6.692 \times 10^{-27}$ - $6.648822 \times 10^{-27}$ = $4.32 \times 10^{-29}$ kg [½]

Accept $4.3 \times 10^{-29}$ kg or similar rounding. Deduct ½ mark for arithmetic error with correct method.

(b) [2 marks]

• The mass defect represents mass that has been converted to energy [1]
• According to Einstein's equation $E = mc^2$ [½], where $c = 3 \times 10^8$ m/s
• A very small mass defect produces a very large amount of energy because $c^2$ is enormous [½]
• This energy is released mainly as kinetic energy of the products and as gamma radiation [½]

Maximum 2 marks.

END OF ANSWER KEY

Total Marks: 40