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प्रश्न
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.)
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उत्तर
When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m
∴ d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C
Potential energy of the two-deuteron system:
`"V" = "e"^2/(4pi in_0 "d")`
Where,
`in_0` = Permittivity of free space
`1/(4piin_0) = 9 xx 10^9 "N" "m"^2"C"^(-2)`
`therefore "V" = (9 xx 10^9 xx (1.6 xx 10^(-19))^2)/(4 xx 10^(-15)) "J"`
`= (9 xx 10^9 xx (1.6 xx 10^(-19))^2)/(4 xx 10^(-15) xx (1.6 xx 10^(-19)) " eV"`
= 360 keV
Hence, the height of the potential barrier of the two-deuteron system is 360 keV.
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
How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
\[\ce{^2_1H + ^2_1H -> ^3_1He + n + 3.27 MeV}\]
Write notes on Nuclear fission
Write notes on Nuclear fusion
Write one balanced equation to show Nuclear fission
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.)
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