Consider the D–T reaction (deuterium–tritium fusion)2 3 41 1 2 H H He n + → +(a) Calculate the energy released in MeV in this reaction from thedata:m(21H )=2.014102 um(31H ) =3.016049 u(b) Consider the radius of both deuterium and tritium to beapproximately 2.0 fm. What is the kinetic energy needed toovercome the coulomb repulsion between the two nuclei? To whattemperature must the gas be heated to initiate the reaction?(Hint: Kinetic energy required for one fusion event =averagethermal kinetic energy available with the interacting particles= 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
a)
Given: the mass of H 1 2 is 2.014102 u , the mass of the H 1 3 is 3.016049 u and the D-T reaction (deuterium-tritium fusion) is H 1 2 + H 1 3 → H 2 4 e + n .
The Q value of the D-T reaction is given as,
Q = ( m 1 + m 2 − m 3 − m 4 ) c 2
Where, mass of H 1 2 is m 1 , the mass of the H 1 3 is m 2 , the mass of H 2 4 e is m 3 and the mass of the n 0 1 is m 4 .
Since, 1 u = 931.5 MeV/c 2 .
By substituting the given values in the above expression, we get
Q = ( 2.014102 + 3.016049 − 4.002603 − 1.008665 ) c 2 = ( 0.018883 c 2 ) u = 0.018883 × 931.5 = 17.59 MeV
Thus, the Q value of fission process is 17.59 MeV .
b)
Given: The radius of deuterium and tritium is 2.0 fm .
Since, the distance between the two nuclei when they touch each other will be the sum of the radius of both nuclei.
The repulsive potential energy between two nuclei is given as,
V = e 2 4 π ε ∘ ( d )
Where, the permittivity of free space is ε ∘ , the distance between the two nuclei d , the charge on the deuterium and tritium nucleus is e and the repulsive potential energy is V .
Since, 1 fm = 1 × 10 − 15 m .
By substituting the given values in the above formula we get.
V = ( 1.6 × 10 − 19 ) 2 4 π × 8.85 × 10 − 12 × 4 × 10 − 15 = 9 × 10 9 × ( 1.6 × 10 − 19 ) 2 4 × 10 − 15 = 5.76 × 10 − 14 J
The temperature required for triggering the reaction is given as,
T = V 3 k
Where, Boltzmann constant is k and the temperature required for triggering the reaction is T
By substituting the given values in the above formula, we get
T = 5.76 × 10 − 14 3 × 1.38 × 10 − 23 = 1.39 × 10 9 K
Thus, to initiate the reaction the gas has to be heated at temperature of 1.39 × 10 9 K and the needed kinetic energy is 5.76 × 10 − 14 J .
a)
Given: the mass of
The
Where, mass of
Since,
By substituting the given values in the above expression, we get
Thus, the
b)
Given: The radius of deuterium and tritium is
Since, the distance between the two nuclei when they touch each other will be the sum of the radius of both nuclei.
The repulsive potential energy between two nuclei is given as,
Where, the permittivity of free space is
Since,
By substituting the given values in the above formula we get.
The temperature required for triggering the reaction is given as,
Where, Boltzmann constant is
By substituting the given values in the above formula, we get
Thus, to initiate the reaction the gas has to be heated at temperature of
