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Secondary 4 Pure Physics Modern Physics Quiz

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Questions

Secondary 4 Pure Physics Quiz - Modern Physics

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly; no marks will be given for correct answers without working.
The use of an approved scientific calculator is expected.
Where appropriate, take the acceleration of free fall, $g = 10 \text{ m/s}^2$ .

Section A: Multiple Choice Questions (Questions 1–10)

Each question carries 1 mark.

1. Which of the following correctly describes the structure of a neutral atom of Carbon-14 ( $^{14}_{6}\text{C}$ )?

A. 6 protons, 8 neutrons, 6 electrons
B. 6 protons, 6 neutrons, 8 electrons
C. 8 protons, 6 neutrons, 6 electrons
D. 6 protons, 8 neutrons, 8 electrons

Answer space

2. A radioactive source emits radiation that is stopped by a few millimetres of aluminium but passes through several centimetres of air. What type of radiation is this?

A. Alpha particles
B. Beta particles
C. Gamma rays
D. Neutrons

Answer space

3. Which statement about $\alpha$ -particles, $\beta$ -particles, and $\gamma$ -rays is correct?

A. $\alpha$ -particles are the most penetrating.
B. $\beta$ -particles are negatively charged electrons.
C. $\gamma$ -rays are deflected most by a magnetic field.
D. $\alpha$ -particles have the highest speed in a vacuum.

Answer space

4. The half-life of a radioactive isotope is 12 hours. If a sample initially has an activity of 800 counts per minute, what will be its activity after 2 days?

A. 50 counts per minute
B. 100 counts per minute
C. 200 counts per minute
D. 400 counts per minute

Answer space

5. In a nuclear fission reaction, a uranium-235 nucleus absorbs a neutron and splits into two smaller nuclei and releases energy. What is the primary source of this released energy?

A. Chemical bonds breaking
B. Conversion of mass into energy
C. Kinetic energy of the incoming neutron
D. Potential energy of the electrons

Answer space

6. Which region of the electromagnetic spectrum is used for sterilising medical equipment due to its high ionising power?

A. Infrared
B. Microwaves
C. Ultraviolet
D. Gamma rays

Answer space

7. An isotope of iodine, $^{131}_{53}\text{I}$ , undergoes beta-minus decay. What is the proton number and nucleon number of the resulting daughter nucleus?

A. Proton number 52, Nucleon number 131
B. Proton number 53, Nucleon number 130
C. Proton number 54, Nucleon number 131
D. Proton number 54, Nucleon number 132

Answer space

8. Why is background radiation subtracted when measuring the activity of a radioactive source in a laboratory?

A. To increase the precision of the Geiger-Müller tube.
B. To ensure the reading reflects only the source’s emission.
C. Because background radiation cancels out beta particles.
D. To calibrate the timer used for counting.

Answer space

9. Which of the following is a characteristic of nuclear fusion?

A. It occurs at room temperature and pressure.
B. It involves the splitting of heavy nuclei.
C. It produces radioactive waste with long half-lives.
D. It releases energy when light nuclei combine to form a heavier nucleus.

Answer space

10. A student plots a graph of count rate against time for a radioactive source. The curve does not approach zero but levels off at 20 counts/min. What does this value of 20 counts/min represent?

A. The initial activity of the source.
B. The half-life of the source.
C. The background radiation count rate.
D. The maximum detection limit of the instrument.

Answer space

Section B: Structured Questions (Questions 11–17)

Questions carry marks as indicated.

11. The diagram below represents the nucleus of an atom of Lithium-7 ( $^{7}_{3}\text{Li}$ ).

Image pending generation for this question.

(a) State the number of protons and neutrons in this nucleus. [2]
Protons: _______________
Neutrons: _______________

(b) Define the term isotope. [2]

Answer space

12. A radioactive source is placed in front of a detector. Sheets of different materials are placed between the source and the detector. The results are shown below:

Material	Thickness	Count Rate (counts/min)
None	-	500
Paper	0.1 mm	480
Aluminium	3 mm	480
Lead	5 mm	50

(a) Identify the type(s) of radiation emitted by the source. Explain your answer. [3]

(b) Suggest why the count rate did not drop to zero even after placing the lead sheet. [1]

Answer space

13. Cobalt-60 is a radioactive isotope used in radiotherapy to treat cancer. It has a half-life of 5.3 years.

(a) Define half-life. [2]

(b) A hospital receives a fresh source of Cobalt-60 with an activity of 4000 Bq. Calculate the activity of the source after 10.6 years. [2]

Answer space

(c) Explain why Cobalt-60 is suitable for treating deep-seated tumours, referring to the properties of the radiation it emits. [2]

Answer space

14. In a nuclear power station, Uranium-235 undergoes fission.

(a) Complete the equation for the fission of Uranium-235: [2]
$^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + \dots ^{1}_{0}\text{n}$
Number of neutrons released: _______________

(b) Describe the role of the moderator in a nuclear reactor. [2]

Answer space

15. The electromagnetic spectrum is shown below in order of increasing frequency.

Radio Waves $\rightarrow$ Microwaves $\rightarrow$ Infrared $\rightarrow$ Visible Light $\rightarrow$ Ultraviolet $\rightarrow$ X-rays $\rightarrow$ Gamma Rays

(a) State two properties common to all electromagnetic waves in a vacuum. [2]

(b) Give one use for Microwaves and one hazard associated with excessive exposure to Ultraviolet radiation. [2]
Use of Microwaves: ______________________________________________________
Hazard of UV: __________________________________________________________

Answer space

16. A Geiger-Müller tube is connected to a counter to measure the activity of a source.

(a) Explain why the source should be handled with tongs rather than by hand. [1]

(b) The background count rate is 20 counts/min. With the source present, the count rate is 220 counts/min.
Calculate the corrected count rate due to the source alone. [1]

Answer space

(c) If the source is moved further away from the detector, the count rate decreases. Explain why, in terms of the geometry of radiation propagation. [2]

Answer space

17. Consider the fusion reaction that powers the Sun:
$^{2}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow ^{4}_{2}\text{He} + ^{1}_{0}\text{n} + \text{Energy}$

(a) Why is this reaction difficult to achieve on Earth? [2]

(b) Despite the difficulties, why is fusion considered a desirable future energy source? [2]

Answer space

Section C: Free Response Question (Question 18-20 Context)

This section tests application and synthesis.

18. A scientist is investigating a new radioactive isotope, X. She measures the count rate every 10 minutes. The background radiation is 10 counts/min. Her results are shown below:

Time (min)	Measured Count Rate (counts/min)
0	810
10	410
20	210
30	110
40	60

(a) Complete the table by calculating the Corrected Count Rate for each time interval. [2]

Time (min)	Corrected Count Rate (counts/min)
0	_______________
10	_______________
20	_______________
30	_______________
40	_______________

(b) Plot a graph of Corrected Count Rate (y-axis) against Time (x-axis) on the grid provided (or sketch the shape). [2]
Describe the shape of the graph.

Answer space

19. Using the half-life calculated in Q18(c), predict the corrected count rate after 50 minutes. [2]

Answer space

20. Isotope X is found to emit beta particles.
(a) Describe the change in the nucleus when a beta particle is emitted. [2]

(b) Why is a beta-emitting source unsuitable for sterilising sealed surgical instruments inside thick metal containers? [2]

Answer space

End of Quiz

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Answers

Secondary 4 Pure Physics Quiz - Modern Physics (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions

1. A
Reasoning: Carbon-14 ( $^{14}_{6}\text{C}$ ) has proton number 6 (so 6 protons and 6 electrons in a neutral atom). Nucleon number is 14, so neutrons = $14 - 6 = 8$ .

2. B
Reasoning: Alpha is stopped by paper. Beta is stopped by a few mm of aluminium. Gamma is only significantly reduced by thick lead. Since it passes through air but is stopped by Al, it is Beta.

3. B
Reasoning: Beta particles are high-speed electrons (negative charge). Alpha are least penetrating. Gamma are uncharged (not deflected). Alpha are slowest/heaviest.

4. A
Reasoning: 2 days = 48 hours. Half-life = 12 hours. Number of half-lives = $48/12 = 4$ .
Activity = $800 \times (1/2)^4 = 800 / 16 = 50$ counts/min.

5. B
Reasoning: In nuclear reactions, a small amount of mass is converted into a large amount of energy according to $E=mc^2$ .

6. D
Reasoning: Gamma rays have high penetration and ionising power, suitable for sterilising sealed equipment. UV is used for surface sterilisation but Gamma is standard for medical packs.

7. C
Reasoning: Beta-minus decay: a neutron turns into a proton and an electron. Proton number increases by 1 ( $53 \rightarrow 54$ ). Nucleon number stays the same (131).

8. B
Reasoning: Background radiation is always present. To find the activity of the source specifically, background counts must be subtracted.

9. D
Reasoning: Fusion is the combining of light nuclei (e.g., Hydrogen isotopes) to form a heavier nucleus (Helium), releasing energy.

10. C
Reasoning: Radioactive decay approaches zero asymptotically. If it levels off at a non-zero value, that value is the constant background radiation.

Section B: Structured Questions

11.
(a) Protons: 3 [1], Neutrons: 4 [1] ( $7-3=4$ )
(b) Isotopes are atoms of the same element (same proton number) [1] with different numbers of neutrons (different nucleon number) [1].
(c) Atoms of the same element have the same number of protons [1] but can have different numbers of neutrons, leading to different total masses [1].

12.
(a) Alpha and Gamma [1 for identification, 1 for explanation].
Explanation: The drop from 500 to 480 with paper is small, suggesting Alpha is mostly blocked or absent? Wait. Alpha is stopped by paper. If it were pure Alpha, it would drop to background. The small drop suggests Alpha might be present but weak, or mostly Beta/Gamma. However, Aluminium (3mm) stops Beta. The count stays at 480 after Al, meaning no Beta was present (or negligible). The drop to 50 with Lead indicates Gamma is present.
Correction/Refinement based on standard patterns:

Paper stops Alpha. Count drops slightly (500->480). This implies very little Alpha or experimental variance, OR the source is Beta/Gamma.
Aluminium stops Beta. Count remains 480. This implies no Beta radiation.
Lead reduces count to 50. This implies Gamma radiation is present (partially absorbed).
The remaining 50 is likely background.
Re-evaluating standard exam logic: Usually, a drop with paper indicates Alpha. A drop with Al indicates Beta. A drop with Pb indicates Gamma.
Here: Paper (0.1mm) -> 480. (Small drop). Al (3mm) -> 480. (No change). Pb (5mm) -> 50. (Large drop).
Conclusion: The source emits Gamma rays [1]. The small drop with paper might be due to absorption of low-energy Gamma or experimental error, but the lack of change with Aluminium rules out Beta. The significant reduction with Lead confirms Gamma.
Alternative Interpretation: If the question implies standard "stops", then:
If Alpha were present, count would drop significantly with paper. It didn't.
If Beta were present, count would drop with Al. It didn't.
Therefore, it is Gamma only. [1]
Wait, let's look at the counts. 500 to 480 is a 4% drop. 480 to 50 is a ~90% drop.
Answer: Gamma rays [1]. Reason: They are not stopped by paper or aluminium but are partially absorbed by lead [2].
(Note: If the student argues Alpha is present due to the small drop, they must explain why it didn't disappear. Standard answer is Gamma).

(b) Background radiation is always present / The lead does not absorb 100% of the gamma rays [1].

13.
(a) Half-life is the time taken for the activity (or count rate/number of nuclei) of a radioactive source to decrease to half its original value [2].
(b) 10.6 years is exactly 2 half-lives ( $10.6 / 5.3 = 2$ ).
Activity = $4000 \rightarrow 2000 \rightarrow 1000$ Bq.
Answer: 1000 Bq [2].
(c) Gamma rays are highly penetrating [1], allowing them to reach deep-seated tumours inside the body without invasive surgery [1].

14.
(a) Conservation of nucleon number: $235 + 1 = 141 + 92 + x(1)$ .
$236 = 233 + x$ .
$x = 3$ .
Answer: 3 neutrons [2].
(b) The moderator (e.g., graphite or heavy water) slows down the fast neutrons produced by fission [1] so that they can be effectively absorbed by Uranium-235 nuclei to sustain the chain reaction [1].
(c) Does not produce greenhouse gases (CO2) / High energy density / Fuel is abundant [1].

15.
(a) 1. They all travel at the same speed ( $3 \times 10^8$ m/s) in a vacuum [1].
2. They are all transverse waves / They can travel through a vacuum [1].
(b) Use of Microwaves: Satellite communications / Radar / Cooking [1].
Hazard of UV: Skin cancer / Blindness (cataracts) / Sunburn [1].

16.
(a) To minimise radiation exposure to the body/hands [1].
(b) Corrected count rate = $220 - 20 =$ 200 counts/min [1].
(c) Radiation spreads out in all directions (spherically) [1]. As distance increases, the same amount of radiation is spread over a larger area, so fewer particles enter the detector [1].

17.
(a) High temperatures and pressures are required to overcome the electrostatic repulsion between the positively charged nuclei [2].
(b) Fuel (hydrogen isotopes) is abundant/readily available from seawater [1]; Produces no long-lived radioactive waste / No greenhouse gases [1].

Section C: Free Response

18.
(a) Subtract 10 from each measured rate:
0 min: 800
10 min: 400
20 min: 200
30 min: 100
40 min: 50
[2 marks: 1 for all correct, or 1 for 3-4 correct]

(b) Graph: Y-axis (Corrected Count Rate), X-axis (Time). Points plotted correctly. Smooth curve drawn showing exponential decay. [2]
Shape: Exponential decay / Curve decreasing with decreasing gradient [1].

19.
After 50 minutes (5 half-lives):
Current at 40 mins is 50.
One more half-life (10 mins) $\rightarrow 50 / 2 =$ 25 counts/min [2].

20.
(a) A neutron changes into a proton and an electron (beta particle) [1]. The proton number increases by 1, nucleon number remains unchanged [1].
(b) Beta particles have low penetrating power (stopped by a few mm of aluminium/metal) [1]. They would not penetrate the thick metal container to reach the instruments inside [1].