Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy released in MeV in the deuterium-tritium fusion reaction $^{2}_{1}\mathrm{H} +\ ^{3}_{1}\mathrm{H} \rightarrow\ ^{4}_{2}\mathrm{He} + n$ using the data $m(^{2}_{1}\mathrm{H}) = 2.014102\ u$, $m(^{3}_{1}\mathrm{H}) = 3.016049\ u$, $m(^{4}_{2}\mathrm{He}) = 4.0026033\ u$, $m_n = 1.008665\ u$, $1\ u = 931.5\ \mathrm{MeV}/c^2$.

CBSE Class 12 · 2015 · Delhi · Modern PhysicsShort Answer

Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy released in MeV in the deuterium-tritium fusion reaction

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n

using the data

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 3.016049 u

m (_{2}^{4} He) = 4.0026033 u

m_{n} = 1.008665 u

1 u = 931.5 MeV / c^{2}

Answer: Fission splits / fusion combines (both raise BE/A); $Q \approx 17.6 MeV$ .

Solution

Fission: a heavy nucleus splits into two lighter nuclei; fusion: two light nuclei combine into a heavier nucleus. In both, the products have a higher binding energy per nucleon than the reactants, so mass is converted into energy (

E = Δ m c^{2}

).
Mass defect of the D–T reaction:

Δ m = (2.014102 + 3.016049) - (4.0026033 + 1.008665) = 5.030151 - 5.011268 = 0.018883 u .

Q = 0.018883 \times 931.5 \approx 17.6 MeV .

Q \approx 17.6 MeV

CBSE Class 12 Board · 2015 · Delhi · Subjective

Practise more: all CBSE board PYQs on Modern Physics →

CBSE Class 12 · 2015 · Delhi · Modern PhysicsShort Answer

Solution

More board questions on Modern Physics