The binding energy per nucleon for deuteron (1H2) and helium (2He4) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus is

Question

The binding energy per nucleon for deuteron (1H2) and helium (2He4) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus is (A) 36.2 MeV (B) 23.6 MeV (C) 47.2 MeV (D) 11.8 MeV

Tardigrade Admin · Accepted Answer

The nuclear reaction is, 1 H 2+ 1 H 2 longrightarrow 2 He 4+Q Total binding energy of helium nucleus =4 × 7=28 MeV Total binding energy of each deuteron =2 × 1.1=2.2 MeV Hence, energy released when two deuteron fuse to form helium, =28-2 × 2.2=28-4.4=23.6 MeV

Q. The binding energy per nucleon for deuteron $(_{1} H^{2})$ and helium $(_{2} H e^{4})$ are $1.1 M e V$ and $7.0 M e V$ respectively. The energy released when two deuterons fuse to form a helium nucleus is

$36.2 M e V$

$23.6 M e V$

$47.2 M e V$

$11.8 M e V$

$9.31 M e V$

Q. The binding energy per nucleon for deuteron (1​H2) and helium (2​He4) are 1.1MeV and 7.0MeV respectively. The energy released when two deuterons fuse to form a helium nucleus is

36.2MeV

23.6MeV

47.2MeV

11.8MeV

9.31MeV