JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 26)
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
30.2 MeV
32.4 MeV
23.6 MeV
25.8 MeV
Explanation
1H2 + 1H2 $$ \to $$ 2He4
No. of proton in one dueteron = 2
$$\therefore\,\,\,$$ Total protons in two dueterons = 2 $$ \times $$ 2 = 4
$$\therefore\,\,\,$$ Binding energy of two dueteron
= 1.1 $$ \times $$ 4 = 4 : 4 MeV
In (2He4) no of protons = 4
$$\therefore\,\,\,$$ Binding energy of (2He4) nuclei = 4 $$ \times $$ 7 = 28 Mev
Energy released in this process = 28 $$-$$ 4.4 = 23.6 MeV
No. of proton in one dueteron = 2
$$\therefore\,\,\,$$ Total protons in two dueterons = 2 $$ \times $$ 2 = 4
$$\therefore\,\,\,$$ Binding energy of two dueteron
= 1.1 $$ \times $$ 4 = 4 : 4 MeV
In (2He4) no of protons = 4
$$\therefore\,\,\,$$ Binding energy of (2He4) nuclei = 4 $$ \times $$ 7 = 28 Mev
Energy released in this process = 28 $$-$$ 4.4 = 23.6 MeV
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