Back to Course

SS3: PHYSICS - 2ND TERM

0% Complete
0/0 Steps
  1. Magnetic Field
    10 Topics
    |
    1 Quiz
  2. Magnetic Field
    5 Topics
    |
    1 Quiz
  3. Electromagnetic Field
    4 Topics
    |
    1 Quiz
  4. Electromagnetic Induction
    7 Topics
    |
    1 Quiz
  5. The Transformer
    5 Topics
    |
    1 Quiz
  6. Simple A.C Circuit
    11 Topics
    |
    1 Quiz
  7. Models of the Atom
    2 Topics
    |
    1 Quiz
  8. Radioactivity
    3 Topics
    |
    1 Quiz
  9. Half Life
    8 Topics
    |
    1 Quiz
  10. Energy Quantization
    5 Topics
    |
    1 Quiz
  11. Photoelectric Effect
    6 Topics
    |
    1 Quiz
  12. Wave Particle Paradox
    6 Topics
    |
    1 Quiz
  • excellence
  • Follow

Lesson 9, Topic 4
In Progress

Nuclear Energy

Lesson Progress
0% Complete

Topic Content:

  • 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.

Nuclear energy-binding energy

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 vacuum = 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 Joules 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

avatar
Subscribe
Notify of
guest
0 Comments
Oldest
Newest
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x