13.28 Consider the D–T reaction (deuterium–tritium fusion)

H + H H e + n 2 4 1 3 1 2

(a) Calculate the energy released in MeV in this reaction from the data:

m( H ) 1 2  =2.014102 u

m( H ) 1 3  =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

(Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmann's constant, T = absolute temperature.)

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7 months ago

13.28 The equation for deuterium-tritium fusion is given as:

H + H H e + n 2 4 1 3 1 2

 It is given that

Mass of ( H ) , m 1 1 2  = 2.014102 u

Mass of ( H ) 1 3 , m 2 = 3.016049 u

Mass of ( H e ) , m 3 2 4  = 4.002603 u

Mass of ( n ) , m 4 0 1 = 1.008665 u

Q-value of the given D-T reaction is:

Q = m 1 + m 2 - m 3 - m 4 c 2

= 2.014102 + 3 . 016049 - 4.002603 - 1.008665 c 2 u

= 0.018883 c 2 u

= 17.59 MeV

Radius of the deuterium and tritium, r 2.0 f m = 2 * 10 - 15 m

Distance

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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