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Additional Exercises · 13.28

Q.Consider the D-T reaction (deuterium-tritium fusion)
[!FORMULA] 12H+13H→24He+n^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + n

(a) Calculate the energy released in MeV in this reaction from the data:
m(12H)=2.014102m(^{2}_{1}\text{H}) = 2.014102 u
m(13H)=3.016049m(^{3}_{1}\text{H}) = 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)2(3kT/2); kk = Boltzmann's constant, TT = absolute temperature.)
Rajasthan RbseTextbookSubjective· 3mImportance★★★★★
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  1. Mass-defect calculation gives Q≈17.6Q\approx17.6 MeV for D-T fusion.
  2. The Coulomb barrier at 4 fm separation is ≈0.36\approx0.36 MeV, and equating this to the average thermal kinetic energy 3kT3kT gives an ignition temperature of ≈1.39×109\approx1.39\times10^9 K — illustrating why practical fusion needs enormous confined temperatures (real reactors get away with somewhat lower temperatures thanks to quantum tunneling and the high-energy tail of the thermal distribution, but the order of magnitude here is the right one).
  1. Energy released

    Q=[m(2H)+m(3H)−m(4He)−mn]×931.5 MeVQ = \left[m(^2\text{H}) + m(^3\text{H}) - m(^4\text{He}) - m_n\right] \times 931.5\ \text{MeV}

    Using m(4He)=4.002603m(^4\text{He})=4.002603 u and mn=1.008665m_n=1.008665 u (given earlier in this chapter's data):

    reactants=2.014102+3.016049=5.030151 u\text{reactants} = 2.014102 + 3.016049 = 5.030151\ \text{u}

    products=4.002603+1.008665=5.011268 u\text{products} = 4.002603 + 1.008665 = 5.011268\ \text{u}

    Δm=5.030151−5.011268=0.018883 u\Delta m = 5.030151 - 5.011268 = 0.018883\ \text{u}

    Q=0.018883×931.5=17.59 MeVQ = 0.018883 \times 931.5 = 17.59\ \text{MeV}

  2. Coulomb barrier and ignition temperature At the moment the two nuclei just touch, the centre-to-centre separation is:

    d=rD+rT=2.0+2.0=4.0 fm=4.0×10−15 md = r_D + r_T = 2.0 + 2.0 = 4.0\ \text{fm} = 4.0 \times 10^{-15}\ \text{m}

    The Coulomb potential energy at this separation (both nuclei carry charge +e+e): V=14πε0e2d=(9×109)(1.6×10−19)24.0×10−15V = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{d} = (9\times 10^9)\frac{(1.6\times 10^{-19})^2}{4.0\times 10^{-15}} …

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