|Class 12 Physics Nuclei||Nuclear Fusion|
Energy generation in sun
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
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).
Answer:- We have the expression for nuclear radius as:
R = R0A1/3
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
Hence, the nuclear matter density is independent of A. It is nearly constant.