Two nuclei of deuterium and one nucleus of tritium combine to form one nucleus of…

PHYSICS · Multi-part · CBSE · Physics · Nuclei · Nuclei

Problem

Two nuclei of deuterium

(^2_1\mathrm D)

and one nucleus of tritium

(^3_1\mathrm T)

combine to form one nucleus of helium-4

(^4_2\mathrm{He})

and one free neutron according to the reaction

^2_1\mathrm D + ^3_1\mathrm T \rightarrow ^4_2\mathrm{He} + ^1_0\mathrm n.

The following atomic masses are provided:

^2_1\mathrm D = 2.014102\,\text{u}

^3_1\mathrm T = 3.016049\,\text{u}

^4_2\mathrm{He} = 4.002603\,\text{u}

^1_0\mathrm n = 1.008665\,\text{u}

. Take

1\,\text{u} = 931.5\,\text{MeV}\,c^{-2}

. (i) Calculate the mass defect for one complete fusion reaction. (ii) Hence determine the energy released per reaction in MeV. (iii) Estimate the energy, in joules, obtained when

1\,\text{kg}

of a 50-50 (by number of nuclei) mixture of deuterium and tritium undergoes this fusion reaction completely. (iv) The fission of one

^{235}\mathrm U

nucleus releases about 200 MeV. For the same number of reactions, by what factor is the energy released in the above fusion process greater or smaller than that from

^{235}\mathrm U

fission?

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