The binding energy per nucleon of deuteron $(_{1} H^{2})$ and helium nucleus $(_{2} {H e}^{4})$ is $1.1$ $M e V$ and $7$ $M e V$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is
$13.9 M e V$
$19.2 M e V$
$26.9 M e V$
$23.6 M e V$

Question

The binding energy per nucleon of deuteron $(_{1} H^{2})$ and helium nucleus $(_{2} {H e}^{4})$ is $1.1$ $M e V$ and $7$ $M e V$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is
$13.9 M e V$
$19.2 M e V$
$26.9 M e V$
$23.6 M e V$

Toppr Admin · Accepted Answer

Energy required to break 4 nucleons in the two deuterons -4-xD7-1-1-xA0-MeVEnergy released in formation of 4 nucleons in the single helium nucleus -4-xD7-7-xA0-MeVHence- total energy released -28-x2212-4-4-MeV-23-6-xA0-MeV

The binding energy per nucleon of deuteron (1H2) and helium nucleus (2He4) is 1.1MeV and 7MeV respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is13.9MeV19.2MeV26.9MeV23.6MeV

Energy required to break 4 nucleons in the two deuterons =4×1.1 MeVEnergy released in formation of 4 nucleons in the single helium nucleus =4×7 MeVHence, total energy released =(28−4.4)MeV=23.6 MeV

The binding energy per nucleon of deuteron $(_{1} H^{2})$ and helium nucleus $(_{2} {H e}^{4})$ is $1.1$ $M e V$ and $7$ $M e V$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is
$13.9 M e V$
$19.2 M e V$
$26.9 M e V$
$23.6 M e V$

Energy required to break $4$ nucleons in the two deuterons $= 4 \times 1.1 M e V$
Energy released in formation of $4$ nucleons in the single helium nucleus $= 4 \times 7 M e V$
Hence, total energy released $= (28 - 4.4) M e V = 23.6 M e V$