Edexcel A Level (IAL) Physics-5.10 Atomic Mass Unit- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -5.10 Atomic Mass Unit- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -5.10 Atomic Mass Unit- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Using the Atomic Mass Unit (u) and Converting to SI Units
The atomic mass unit (u) is a convenient unit for expressing the extremely small masses of atoms, nuclei, and subatomic particles.
Definition of the Atomic Mass Unit
The atomic mass unit is defined as:
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\( 1\ \mathrm{u} = \dfrac{1}{12} \times \text{mass of a carbon-12 atom} \)
In SI units:
\( 1\ \mathrm{u} = 1.66\times10^{-27}\ \mathrm{kg} \)
This small unit makes nuclear and atomic mass values easier to work with.
Why Use the Atomic Mass Unit?
- Masses of atoms and nuclei are extremely small in kilograms.
- Using kilograms leads to inconvenient numbers.
- The unit \( \mathrm{u} \) provides simpler numerical values.
Example: The mass of a proton is approximately \( 1.67\times10^{-27}\ \mathrm{kg} \) or \( \approx 1.01\ \mathrm{u} \).
Converting Between u and kg
(a) From u to kg
Multiply by \( 1.66\times10^{-27} \):
\( m(\mathrm{kg}) = m(\mathrm{u}) \times 1.66\times10^{-27} \)
(b) From kg to u
Divide by \( 1.66\times10^{-27} \):
\( m(\mathrm{u}) = \dfrac{m(\mathrm{kg})}{1.66\times10^{-27}} \)
Typical Atomic and Nuclear Masses
- Proton: \( 1.01\ \mathrm{u} \)
- Neutron: \( 1.01\ \mathrm{u} \)
- Electron: \( 5.49\times10^{-4}\ \mathrm{u} \)
- Carbon-12 nucleus: \( 12.0\ \mathrm{u} \)
Importance in Nuclear Physics
- Used to calculate mass defect
- Used in nuclear binding energy calculations
- Simplifies comparison of nuclear masses
- Essential for using \( \Delta E = c^2 \Delta m \)
Example (Easy)
Convert a mass of \( 2.50\ \mathrm{u} \) into kilograms.
▶️ Answer / Explanation
\( m = 2.50 \times 1.66\times10^{-27} = 4.15\times10^{-27}\ \mathrm{kg} \)
Example (Medium)
The mass of an electron is \( 9.11\times10^{-31}\ \mathrm{kg} \). Convert this mass into atomic mass units.
▶️ Answer / Explanation
\( m = \dfrac{9.11\times10^{-31}}{1.66\times10^{-27}} \)
\( m = 5.49\times10^{-4}\ \mathrm{u} \)
Example (Hard)
A nucleus has a mass defect of \( 0.025\ \mathrm{u} \). Calculate the energy released.
▶️ Answer / Explanation
Convert mass defect to kg:
\( \Delta m = 0.025 \times 1.66\times10^{-27} = 4.15\times10^{-29}\ \mathrm{kg} \)
Calculate energy:
\( \Delta E = c^2 \Delta m = (3.00\times10^8)^2 \times 4.15\times10^{-29} \)
\( \Delta E = 3.74\times10^{-12}\ \mathrm{J} \)
