Q=∑B.E.ofproducts−∑B.E.ofreactants
The equation for two deuterons combining to form Helium nucleus is given as H2+1H2→2H4
Energyreleased=Q=4[B.E.(2H4)]−4[B.E.(1H2)]
=4×7−4×1.1
=28−4.4
The energy released in this process is =23.6MeV
JEE Main 2017 — Physics Modern Physics
Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron=1.1MeV and for helium=7.0MeV)
Held on 8 Apr 2017 · Verified 6 Jul 2026.
23.6MeV
30.2MeV
25.8MeV
32.4MeV
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