QJEEM17Phe52f: JEE Main 2017 Physics - Class 12 JEEM17Phe52f Solution: Two deuterons undergo nuclear fusion to form a Helium nucleus.…

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