Printable worksheet on radioactivity covering topics like nuclear decay, half-life, and properties of radiation particles.
A worksheet titled "Test on Radioactivity" with questions about radioactive sources, half-life, nuclear equations, and properties of alpha and beta particles.
JPG
1000×1291
126.1 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1074989
⭐
Show Answer Key & Explanations
Step-by-step solution for: Chapter Test on Radioactivity worksheet
▼
Show Answer Key & Explanations
Step-by-step solution for: Chapter Test on Radioactivity worksheet
Let's go through each question in the Test on Radioactivity and solve them step by step with clear explanations.
---
#### a. Describe how this could be done using a radioactive source and a counter. *(2 marks)*
Answer:
A radioactive source (emitting radiation such as beta or gamma rays) is placed on one side of the steel sheet, and a detector (counter) is placed on the other side. As the steel moves past the source and detector, the amount of radiation detected will vary depending on the thickness of the steel. Thicker sections absorb more radiation, so fewer counts are recorded; thinner sections allow more radiation through, resulting in higher counts. By monitoring the count rate, any variations in thickness can be detected and corrected.
Explanation:
This method uses the principle that radiation is absorbed by materials. The absorption depends on the density and thickness of the material. So, if the thickness varies, the intensity of radiation reaching the counter changes. This allows for real-time quality control.
---
#### b. Name the type of radiation you would expect the source to emit. *(1 mark)*
Answer:
Beta radiation (or gamma radiation, depending on context)
Explanation:
- Beta radiation is suitable because it can penetrate steel sheets of moderate thickness but is still affected by small changes in thickness.
- Gamma radiation might also be used if the steel is thicker, since gamma rays have greater penetrating power.
- However, beta particles are commonly used for industrial thickness gauging of thin to medium-thickness metals like steel because they are moderately penetrating and sensitive to small thickness changes.
> ✔ Most appropriate answer: Beta radiation
---
#### c. A radioactive source has a half-life of 12 hours. Define the term half-life. *(1 mark)*
Answer:
Half-life is the time taken for half of the radioactive atoms in a sample to decay.
Explanation:
It is a measure of the rate of radioactive decay. After one half-life, 50% of the original nuclei remain; after two half-lives, 25%, and so on. It is constant for a given isotope.
---
| Nuclide | Mass Number | Atomic Number |
|--------|-------------|----------------|
| Cu | 58 | 29 |
| Co | 54 | 27 |
| Cu | 59 | 29 |
| Zn | 58 | 30 |
| Zn | 59 | 30 |
---
#### a. Write a nuclear equation to show a nuclide (daughter nuclide) which could be produced from another nuclide (parent nuclide) by the emission of an alpha particle. *(2 marks)*
Answer:
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{27}^{54}\text{Co} + {}_2^4\text{He}
$$
Explanation:
An alpha particle is a helium nucleus (${}_2^4\text{He}$). When emitted:
- Mass number decreases by 4.
- Atomic number decreases by 2.
So, look for a nuclide where mass number – 4 and atomic number – 2 gives another nuclide in the list.
Check:
- $^{58}_{29}\text{Cu}$ → $^{54}_{27}\text{Co}$? Yes!
- 58 – 4 = 54
- 29 – 2 = 27
→ $^{54}_{27}\text{Co}$ exists in the list.
✔ Valid alpha decay.
---
#### b. Write a nuclear equation to show a nuclide (daughter nuclide) which could be produced from another nuclide (parent nuclide) by the emission of a beta particle. *(2 marks)*
Answer:
$$
{}_{27}^{54}\text{Co} \longrightarrow {}_{28}^{54}\text{Ni} + {}_{-1}^0\beta
$$
But wait — Ni (Nickel) isn't listed. We must use only the given nuclides.
Let’s find a possible beta decay among the given ones.
In beta minus decay, a neutron turns into a proton, emitting an electron (beta particle):
- Mass number stays the same.
- Atomic number increases by 1.
Look at:
- $^{58}_{29}\text{Cu}$ → $^{58}_{30}\text{Zn}$? Yes!
- 29 → 30, mass 58 → 58 → matches!
So:
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{30}^{58}\text{Zn} + {}_{-1}^0\beta
$$
✔ This is valid and uses only given nuclides.
---
#### c. Name the nuclide which possesses the most neutrons. *(1 mark)*
Answer:
$^{59}_{30}\text{Zn}$ (Zinc-59)
Explanation:
Number of neutrons = mass number – atomic number.
Calculate for each:
- $^{58}_{29}\text{Cu}$: 58 – 29 = 29 neutrons
- $^{54}_{27}\text{Co}$: 54 – 27 = 27 neutrons
- $^{59}_{29}\text{Cu}$: 59 – 29 = 30 neutrons
- $^{58}_{30}\text{Zn}$: 58 – 30 = 28 neutrons
- $^{59}_{30}\text{Zn}$: 59 – 30 = 29 neutrons
Wait:
- $^{59}_{29}\text{Cu}$ → 30 neutrons
- $^{59}_{30}\text{Zn}$ → 29 neutrons
So $^{59}_{29}\text{Cu}$ has 30 neutrons — highest.
✔ Correct answer: $^{59}_{29}\text{Cu}$ (Copper-59)
> But note: Is Copper-59 stable? Not really, but it's given, so we use it.
So final answer: $^{59}_{29}\text{Cu}$
---
#### a. Explain what those numbers indicate about the nucleus of the atom. *(2 marks)*
Answer:
- The atomic number (90) indicates the number of protons in the nucleus.
- The mass number (232) indicates the total number of protons and neutrons in the nucleus.
So, this thorium atom has:
- 90 protons,
- 232 – 90 = 142 neutrons.
Explanation:
The atomic number defines the element (Th = 90), and the mass number tells us the total nucleons (protons + neutrons).
---
#### b. Thorium decays by alpha particle. State two properties of the alpha particle. *(2 marks)*
Answer:
1. Alpha particles consist of 2 protons and 2 neutrons (same as a helium nucleus).
2. They are highly ionizing but have low penetration power (stopped by paper or skin).
Other acceptable properties:
- Positively charged (+2)
- Relatively large mass
- Travel short distances in air
✔ Any two of these.
---
---
1a. Use a radioactive source and detector on opposite sides of the steel. Variations in count rate indicate thickness variation.
1b. Beta radiation
1c. Half-life is the time for half the radioactive atoms to decay.
---
2a.
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{27}^{54}\text{Co} + {}_2^4\text{He}
$$
2b.
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{30}^{58}\text{Zn} + {}_{-1}^0\beta
$$
2c. $^{59}_{29}\text{Cu}$ (Copper-59)
---
3a.
- Atomic number (90) = number of protons
- Mass number (232) = total protons + neutrons
3b.
1. Consists of 2 protons and 2 neutrons
2. Highly ionizing / Low penetration
---
Let me know if you'd like this formatted as a printable answer sheet!
---
1. A steel manufacturer wishes to check that the thickness of the steel he produces is uniform.
#### a. Describe how this could be done using a radioactive source and a counter. *(2 marks)*
Answer:
A radioactive source (emitting radiation such as beta or gamma rays) is placed on one side of the steel sheet, and a detector (counter) is placed on the other side. As the steel moves past the source and detector, the amount of radiation detected will vary depending on the thickness of the steel. Thicker sections absorb more radiation, so fewer counts are recorded; thinner sections allow more radiation through, resulting in higher counts. By monitoring the count rate, any variations in thickness can be detected and corrected.
Explanation:
This method uses the principle that radiation is absorbed by materials. The absorption depends on the density and thickness of the material. So, if the thickness varies, the intensity of radiation reaching the counter changes. This allows for real-time quality control.
---
#### b. Name the type of radiation you would expect the source to emit. *(1 mark)*
Answer:
Beta radiation (or gamma radiation, depending on context)
Explanation:
- Beta radiation is suitable because it can penetrate steel sheets of moderate thickness but is still affected by small changes in thickness.
- Gamma radiation might also be used if the steel is thicker, since gamma rays have greater penetrating power.
- However, beta particles are commonly used for industrial thickness gauging of thin to medium-thickness metals like steel because they are moderately penetrating and sensitive to small thickness changes.
> ✔ Most appropriate answer: Beta radiation
---
#### c. A radioactive source has a half-life of 12 hours. Define the term half-life. *(1 mark)*
Answer:
Half-life is the time taken for half of the radioactive atoms in a sample to decay.
Explanation:
It is a measure of the rate of radioactive decay. After one half-life, 50% of the original nuclei remain; after two half-lives, 25%, and so on. It is constant for a given isotope.
---
2. The following are five nuclides (nuclei):
| Nuclide | Mass Number | Atomic Number |
|--------|-------------|----------------|
| Cu | 58 | 29 |
| Co | 54 | 27 |
| Cu | 59 | 29 |
| Zn | 58 | 30 |
| Zn | 59 | 30 |
---
#### a. Write a nuclear equation to show a nuclide (daughter nuclide) which could be produced from another nuclide (parent nuclide) by the emission of an alpha particle. *(2 marks)*
Answer:
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{27}^{54}\text{Co} + {}_2^4\text{He}
$$
Explanation:
An alpha particle is a helium nucleus (${}_2^4\text{He}$). When emitted:
- Mass number decreases by 4.
- Atomic number decreases by 2.
So, look for a nuclide where mass number – 4 and atomic number – 2 gives another nuclide in the list.
Check:
- $^{58}_{29}\text{Cu}$ → $^{54}_{27}\text{Co}$? Yes!
- 58 – 4 = 54
- 29 – 2 = 27
→ $^{54}_{27}\text{Co}$ exists in the list.
✔ Valid alpha decay.
---
#### b. Write a nuclear equation to show a nuclide (daughter nuclide) which could be produced from another nuclide (parent nuclide) by the emission of a beta particle. *(2 marks)*
Answer:
$$
{}_{27}^{54}\text{Co} \longrightarrow {}_{28}^{54}\text{Ni} + {}_{-1}^0\beta
$$
But wait — Ni (Nickel) isn't listed. We must use only the given nuclides.
Let’s find a possible beta decay among the given ones.
In beta minus decay, a neutron turns into a proton, emitting an electron (beta particle):
- Mass number stays the same.
- Atomic number increases by 1.
Look at:
- $^{58}_{29}\text{Cu}$ → $^{58}_{30}\text{Zn}$? Yes!
- 29 → 30, mass 58 → 58 → matches!
So:
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{30}^{58}\text{Zn} + {}_{-1}^0\beta
$$
✔ This is valid and uses only given nuclides.
---
#### c. Name the nuclide which possesses the most neutrons. *(1 mark)*
Answer:
$^{59}_{30}\text{Zn}$ (Zinc-59)
Explanation:
Number of neutrons = mass number – atomic number.
Calculate for each:
- $^{58}_{29}\text{Cu}$: 58 – 29 = 29 neutrons
- $^{54}_{27}\text{Co}$: 54 – 27 = 27 neutrons
- $^{59}_{29}\text{Cu}$: 59 – 29 = 30 neutrons
- $^{58}_{30}\text{Zn}$: 58 – 30 = 28 neutrons
- $^{59}_{30}\text{Zn}$: 59 – 30 = 29 neutrons
Wait:
- $^{59}_{29}\text{Cu}$ → 30 neutrons
- $^{59}_{30}\text{Zn}$ → 29 neutrons
So $^{59}_{29}\text{Cu}$ has 30 neutrons — highest.
✔ Correct answer: $^{59}_{29}\text{Cu}$ (Copper-59)
> But note: Is Copper-59 stable? Not really, but it's given, so we use it.
So final answer: $^{59}_{29}\text{Cu}$
---
3. A particular atom of Thorium (Th) has an atomic number of 90 and a mass number of 232.
#### a. Explain what those numbers indicate about the nucleus of the atom. *(2 marks)*
Answer:
- The atomic number (90) indicates the number of protons in the nucleus.
- The mass number (232) indicates the total number of protons and neutrons in the nucleus.
So, this thorium atom has:
- 90 protons,
- 232 – 90 = 142 neutrons.
Explanation:
The atomic number defines the element (Th = 90), and the mass number tells us the total nucleons (protons + neutrons).
---
#### b. Thorium decays by alpha particle. State two properties of the alpha particle. *(2 marks)*
Answer:
1. Alpha particles consist of 2 protons and 2 neutrons (same as a helium nucleus).
2. They are highly ionizing but have low penetration power (stopped by paper or skin).
Other acceptable properties:
- Positively charged (+2)
- Relatively large mass
- Travel short distances in air
✔ Any two of these.
---
✔ Final Answers Summary:
---
1a. Use a radioactive source and detector on opposite sides of the steel. Variations in count rate indicate thickness variation.
1b. Beta radiation
1c. Half-life is the time for half the radioactive atoms to decay.
---
2a.
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{27}^{54}\text{Co} + {}_2^4\text{He}
$$
2b.
$$
{}_{29}^{58}\text{Cu} \longrightarrow {}_{30}^{58}\text{Zn} + {}_{-1}^0\beta
$$
2c. $^{59}_{29}\text{Cu}$ (Copper-59)
---
3a.
- Atomic number (90) = number of protons
- Mass number (232) = total protons + neutrons
3b.
1. Consists of 2 protons and 2 neutrons
2. Highly ionizing / Low penetration
---
Let me know if you'd like this formatted as a printable answer sheet!
Parent Tip: Review the logic above to help your child master the concept of radioactivity worksheet.
