The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV

The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:

Options

(a) 30.2 MeV
(b) 23.6 MeV
(c) 2.2 MeV
(d) 28.0 MeV

Correct Answer:

23.6 MeV

Explanation:

Binding energy of two ₁H² nuclei = 2 (1.1 x 2) = 4.4 meV
Binding energy of one ₂He⁴ nucleus = 4 x 7.0 = 28 MeV
Energy released = 28 – 4.4 = 23.6 MeV

admin:

Related Questions

  1. The de-Broglie wavelength of an electron having 80 eV of energy is nearly
  2. The difference in the variation of resistance with temperature in a metal
  3. The limiting angle of incidence for an optical ray that can be transmitted
  4. In an AC circuit the potential differences across an inductance and resistance joined
  5. Two conducting spheres of radii 5 cm and 10 cm are given a charge of 15 μC each.