Advertisements
Advertisements
प्रश्न
Why nuclear fusion reaction is also called thermo-nuclear reaction?
उत्तर
This is because it takes place only at high temperatures. If the reacting nuclei are at high temperature, then they will have sufficient energy to overcome coulombian force.
APPEARS IN
संबंधित प्रश्न
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :
`""_1^2H+_1^3H->_2^4He+n`
Using the data :
m(`""_1^2H`) = 2.014102 u
m(`""_1^3H`) = 3.016049 u
m(`""_2^4He`) = 4.002603 u
mn = 1.008665 u
1u = 931.5 MeV/c2
Calculate the height of the potential barrier for a head on collision of two deuterons.
(Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Write notes on Nuclear fission
Write one balanced equation to show Nuclear fission
Write one balanced equation to show Nuclear fusion
During a nuclear fission reaction,
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT equals the Coulomb potential energy at 2 fm.
Calculate the Q-values of the following fusion reactions :-
(a) `""_1^2H + ""_1^2H → ""_1^3H + ""_1^1H`
(b) `""_1^2H + ""_1^2H → ""_2^3H + n`
(c) `""_1^2H + ""_1^3H → _2^4H + n`.
Atomic masses are `m(""_1^2H) = 2.014102 "u", m(""_1^3H) = 3.016049 "u", m(""_2^3He) = 3.016029 "u", m(""_2^4He) = 4.002603 "u".`
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
Briefly explain the elementary particles present in nature.
A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the slab is:
(Given latent heat of fusion of ice = 3.36 × 105 J kg−1)