Class 12 Physics Nuclei Nuclear Fusion

Nuclear Fusion

• In a nuclear fusion reaction two lighter nuclei combine to form a relatively heavier nucleus.
• In this process huge amount of energy is also released.
• Temperature at which protons would have enough energy to overcome the coulomb’s barrier is very high.

Thermonuclear fusion

• Increasing the temperature of the material until the particles have enough energy due to their thermal motions alone –to overcome the coulomb barrier.
• For thermonuclear fusion,extreme conditions of temperature and pressure are required.
• Example of Thermonuclear fusion is generation of energy in stars.
• For example: From the sun we get large amount of energy and this energy generated due to the thermonuclear fusion reaction taking place in the sun.

Energy generation in sun

• In sun the energy generation is a multi- step process.There are total of 4 steps involved in the energy generation inside sun.
• Step1:- 11H(proton) + 11H(proton) -->21H(deuteron)+ e+(positron) + ν(neutrino)+0.42MeV
• Step2:- e+(positron) + e-(electron) -->γ(Gamma rays) + 1.02MeV
• Step3:- 21H(deuteron)+ 11H(proton) -->32He(helium) + γ(Gamma rays)+5.49MeV
• Step4:- 32H + 32H -->42He + 11H+11H+12.86MeV
• Step 1,2,and 3 occur twice in the sun and the step 4 occurs only once.
• When all the above 4 reactions are combined together then four hydrogen atoms combine to form a42He atom with a release of 26.7MeV of energy.
• Final reaction is:- 411H +2e- --> 42He +6 γ+2 ν+26.7MeV
• It is also known as proton-proton cycle because this process starts with protons. Problem:- How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as  21H + 21H -->32He+n+3.27MeV

Answer:- The given fusion reaction is:

21H + 21H -->32He+n+3.27MeV

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.

Therefore, 2.0 kg of deuterium contains= ((6.023x1023)/ (2)) x (2000) =6.023x1026atoms

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27

MeV energy is released.

Therefore, total energy per nucleus released in the fusion reaction:

E= (3.27/2) x 6.023x1026 MeV

= (3.27/2) x 6.023x1026x 1.6x10-19x 106

=1.576x1014 J

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

= (1.576x1014)/ (100x60x60x24x365)

= (4.9 x104) years.

Problem:-  From the relation R = R0A1/3, where R0 is a constant and A is the mass number of anucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

R = R0A1/3

Where,

R0 = Constant.

A = Mass number of the nucleus

Nuclear matter density,ρ = (Mass of the nucleus)/ (volume of nucleus)

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

Therefore ρ=(mA)/((4/3)πR3)=(3mA)/(4π)(R0A(1/3))3=(3m)/(4πR03)

Hence, the nuclear matter density is independent of A. It is nearly constant.

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