SS3: PHYSICS - 2ND TERM
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Magnetic Field10 Topics|1 Quiz
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Magnetic Field
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Magnetic & Non-Magnetic Substances
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Soft & Hard Magnetic Substances
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Temporary & Permanent Magnet
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Methods of Making Magnets
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Demagnetization
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Magnetic Properties of Iron & Steel
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Neutral Point | Magnetic Flux Density | Magnetic Potential
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Characteristics of Lines of Force
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Magnetic Screening
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Magnetic Field
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Magnetic Field5 Topics|1 Quiz
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Electromagnetic Field4 Topics|1 Quiz
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Electromagnetic Induction7 Topics|1 Quiz
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The Transformer5 Topics|1 Quiz
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Simple A.C Circuit11 Topics|1 Quiz
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Models of the Atom2 Topics|1 Quiz
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Radioactivity3 Topics|1 Quiz
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Half Life8 Topics|1 Quiz
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Energy Quantization5 Topics|1 Quiz
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Photoelectric Effect6 Topics|1 Quiz
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Wave Particle Paradox6 Topics|1 Quiz
Nuclear Energy
The protons and neutrons in the nucleus of an atom are called nucleons. The energy required to completely separate the protons and the neutrons of an atom is called binding energy. It’s also the energy that would be released if individual protons and neutrons were combined into a single nucleus.
Binding energy is the energy released to join or fuse together the nucleons of the nucleus.

When the nucleus of an atom is separated, the combined mass of protons and neutrons when added together is greater than the mass of the nucleus as a unit
The difference in mass is called Mass Defect. Mass defect is the gain or loss in mass, produced when the protons and neutrons in the nucleus are separated or joined together.
Mass defect = sum of masses of proton and neutron – mass of nucleus
- Δm = mass defect
Einstein showed that mass and energy are related by the equation:
E = mc2
E = Δmc2
- E = energy released
- Δm = is the mass defect
- c = speed or velocity of light
Electron volt is the energy possessed by an electron that is moving under the influence of a potential difference of one volt. It is the unit of energy in nuclear energy.
Value of electron volt (eV):
- 1 eV = 1.6 × 10-19 J
- 1 MeV = 1.602 × 10-13 J
Atomic mass unit ( a.m.u.):
The unit of energy is also expressed in a.m.u.
1 amu = 1.66 × 10-27 Kg = 931 Mev = 1.49 × 10-10 J
Example 9.4.1:
In a nuclear fission process, the mass defect was 2.4 × 10-5 kg. Calculate the energy released. [speed of the light in vacuumA vacuum is an area of space containing little or no matter (solids, liquids, and gases). It is a space with nothing in it, not even air. More = c = 3.0 × 108 ms-1]
Solution
Recall that,
E = mc2
E = 2.4 × 10-5 × (3 × 108)2
E = 2.4 × 10-5 × 9 × 1016
E = 2.16 × 1012 J
Example 9.4.2:
a. Calculate the energy of 2.25 eV
b. What is the value of 2.5 JoulesJoule is the SI (International System of Units) unit of energy and work. It is equal to the amount of work done when a force of 1 newton displaces a mass... More in a.m.u?
Solution
a. 1 eV = 1.6 × 10-19 J
Therefore,
2.25 eV will equal 2.25 × 1.6 × 10-19
= 3.6 × 10-19 Joules
b. 1 amu = 931 Mev = 1.49 × 10-10 J
x amu = \( \frac{2.5}{1.49 \: \times \: 10^{-10}} \scriptsize \: \times \: 1 \)
= 1.67 × 1010 amu