CIE iGCSE Co-ordinated Sciences-P5.1 The nucleus- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P5.1 The nucleus – Study Notes
CIE iGCSE Co-ordinated Sciences-P5.1 The nucleus – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Structure and Charge of the Nucleus
(a) Composition of the Nucleus:
- The nucleus of an atom is made up of two types of particles:
- Protons → positively charged particles.
- Neutrons → neutral particles with no charge.
- Together, protons and neutrons are called nucleons.
(b) Proton Number (Atomic Number):
- The proton number, often written as \( Z \), is the number of protons in the nucleus of an atom.
- It defines the element (e.g., hydrogen has \( Z = 1 \), carbon has \( Z = 6 \)).
- The total nuclear charge is equal to the number of protons, since each proton has charge \( +1 \).
- Therefore, the relative charge on a nucleus = \( +Z \).
(c) Relative Charges of Subatomic Particles:
- Proton: \( +1 \)
- Neutron: \( 0 \)
- Electron: \( -1 \)
(d) Summary Table of Subatomic Particles:
| Particle | Location | Relative Charge | Relative Mass |
|---|---|---|---|
| Proton | Nucleus | +1 | 1 |
| Neutron | Nucleus | 0 | 1 |
| Electron | Shells around nucleus | –1 | 1/1836 |
Example
What is the relative charge of a nucleus of a carbon atom with 6 protons and 6 neutrons?
▶️Answer/Explanation
Step (1) – Count protons:
Carbon has 6 protons, each with charge \( +1 \).
Step (2) – Neutrons:
Neutrons have charge 0, so they do not affect the total charge.
Step (3) – Total nuclear charge:
\( +6 \).
Final Answer:
The relative charge on the carbon nucleus is \( +6 \).
Nuclear Notation and Atomic Structure
(a) Proton Number (Atomic Number, Z):
- The proton number \( Z \) is the number of protons in the nucleus of an atom.
- It determines the identity of the element (e.g., \( Z = 6 \) → Carbon, \( Z = 8 \) → Oxygen).
- The number of electrons in a neutral atom is also equal to \( Z \).
(b) Nucleon Number (Mass Number, A):
- The nucleon number \( A \) is the total number of protons and neutrons in the nucleus.
- Thus, \( A = Z + N \), where \( N \) = number of neutrons.
(c) Number of Neutrons:
- The number of neutrons can be found by:
- \( N = A – Z \).
(d) Nuclide Notation:
- An isotope is represented using the notation:
- \( _Z^A X \), where:
- \( X \) = chemical symbol of the element
- \( A \) = nucleon number (mass number)
- \( Z \) = proton number (atomic number)
Example
Chlorine has two isotopes: \( _{17}^{35}Cl \) and \( _{17}^{37}Cl \). For each isotope, calculate the number of protons, neutrons, and electrons in a neutral atom.
▶️Answer/Explanation
Step (1) – Proton number:
\( Z = 17 \) → both isotopes have 17 protons.
Step (2) – Neutron number:
For \( _{17}^{35}Cl \): \( N = A – Z = 35 – 17 = 18 \).
For \( _{17}^{37}Cl \): \( N = 37 – 17 = 20 \).
Step (3) – Electrons in neutral atom:
Equal to number of protons → 17 electrons for both isotopes.
Final Answer:
– \( _{17}^{35}Cl \): 17 protons, 18 neutrons, 17 electrons.
– \( _{17}^{37}Cl \): 17 protons, 20 neutrons, 17 electrons.
Isotopes
- Atoms of the same element can exist with different numbers of neutrons in the nucleus.
- Such atoms are called isotopes of the element.
- All isotopes of an element have the same proton number (Z) but different nucleon numbers (A).
- Example: \( _6^{12}C \) and \( _6^{14}C \) are both isotopes of carbon.
Radioactive Isotopes:
- Some isotopes are unstable and their nuclei spontaneously break down (decay), emitting radiation.
- These unstable forms are called radioisotopes.
- Example: \( _6^{14}C \) (Carbon-14) is radioactive, while \( _6^{12}C \) is stable.
Key Points:
- Isotopes of the same element have the same chemical properties (since they have the same number of electrons).
- But their physical properties (such as mass, stability, or radioactivity) can differ.
Example
Oxygen has two stable isotopes, \( _8^{16}O \) and \( _8^{18}O \). Calculate the number of neutrons in each isotope.
▶️Answer/Explanation
Step (1) – Recall formula:
\( N = A – Z \).
Step (2) – For \( _8^{16}O \):
\( N = 16 – 8 = 8 \).
Step (3) – For \( _8^{18}O \):
\( N = 18 – 8 = 10 \).
Final Answer:
– \( _8^{16}O \): 8 neutrons (stable).
– \( _8^{18}O \): 10 neutrons (stable).
Nuclear Reactions: Fission and Fusion
(a) Nuclear Fission:
Nuclear fission is the process in which a large, unstable nucleus splits into two (or more) smaller nuclei, releasing energy.
- It is usually triggered when a heavy nucleus (such as \( ^{235}U \) or \( ^{239}Pu \)) absorbs a slow-moving neutron.
- The fission process releases:
- Two or more smaller nuclei (fission fragments)
- Several free neutrons (which may cause further fission → chain reaction)
- A large amount of energy (from the conversion of nuclear mass into energy, according to \( E = mc^2 \)).
- Two or more smaller nuclei (fission fragments)
- Applications: Used in nuclear power stations and atomic bombs.
(b) Nuclear Fusion:
Nuclear fusion is the process in which two light nuclei combine to form a larger, heavier nucleus, releasing enormous energy.
- Example: In the Sun, hydrogen nuclei fuse to form helium:
\( ^1H + ^1H \rightarrow ^2H + \text{energy} \). - Fusion requires extremely high temperatures and pressures to overcome the repulsion between positively charged nuclei.
- Applications: Powers stars and experimental fusion reactors (tokamaks, laser fusion).
| Process | Description | Example | Applications |
|---|---|---|---|
| Nuclear Fission | Splitting of a heavy nucleus into smaller nuclei | \( ^{235}U + n \rightarrow ^{141}Ba + ^{92}Kr + 3n + \text{energy} \) | Nuclear power plants, atomic bombs |
| Nuclear Fusion | Joining of light nuclei to form a heavier nucleus | \( ^2H + ^3H \rightarrow ^4He + n + \text{energy} \) | Stars (Sun), experimental reactors |
Example
In a fission reaction, \( ^{235}U \) absorbs a neutron and splits into \( ^{141}Ba \), \( ^{92}Kr \), and 3 neutrons. Calculate the total number of nucleons before and after the reaction to show conservation of nucleon number.
▶️Answer/Explanation
Step (1) – Before reaction:
\( ^{235}U + ^1n = 235 + 1 = 236 \) nucleons.
Step (2) – After reaction:
\( ^{141}Ba + ^{92}Kr + 3n = 141 + 92 + 3 = 236 \) nucleons.
Step (3) – Conservation:
Total nucleon number is the same before and after → law of conservation of nucleon number holds.
Final Answer:
236 nucleons before, 236 nucleons after → nucleon number conserved.