The average binding energies of deuterium and $α$ -particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction $_{1} H^{2}$ + $_{1} H^{2}$ $\to$ $_{2} H e^{4}$ is $- ------- -$ MeV.

Question

The average binding energies of deuterium and

α

-particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction

_{1} H^{2}

+

_{1} H^{2}

\to

_{2} H e^{4}

is

- ------- -

MeV.

Toppr Admin · Accepted Answer

Average binding energy of deuterium-xA0-1H2-xA0-xA0-per nucleon -xA0- Ed-1-12 MeVAverage binding energy of -x3B1-particle-xA0-2H4-xA0-xA0-per nucleon -xA0- E-x3B1-7-1 MeVEnergy liberated in the given reaction -xA0-Q-4-E-x3B1-x2212-2-Ed-x2212-2-Ed-x2234- Q-4-7-1-x2212-2-1-12-x2212-2-1-12-23-92 MeV

The average binding energies of deuterium and $α$ -particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction $_{1} H^{2}$ + $_{1} H^{2}$ $\to$ $_{2} H e^{4}$ is $- ------- -$ MeV.
$23.92$
$25$
$25.6$
$15.8$

Average binding energy of deuterium $_{1} H^{2}$ per nucleon $E_{d} = 1.12$ MeV
Average binding energy of $α$ -particle $_{2} H^{4}$ per nucleon $E_{α} = 7.1$ MeV
Energy liberated in the given reaction $Q = 4 (E_{α}) - 2 (E_{d}) - 2 (E_{d})$
$∴$ $Q = 4 (7.1) - 2 (1.12) - 2 (1.12) = 23.92$ MeV

The average binding energies of deuterium and α -particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction 1H2+1H2 → 2He4 is ––––––––– MeV.23.922525.615.8

Average binding energy of deuterium 1H2 per nucleon Ed=1.12 MeVAverage binding energy of α-particle 2H4 per nucleon Eα=7.1 MeVEnergy liberated in the given reaction Q=4(Eα)−2(Ed)−2(Ed)∴ Q=4(7.1)−2(1.12)−2(1.12)=23.92 MeV

The average binding energies of deuterium and $α$ -particle are 1.12 MeV and 7.1MeV respectively. The energy that liberates in the reaction $_{1} H^{2}$ + $_{1} H^{2}$ $\to$ $_{2} H e^{4}$ is $- ------- -$ MeV.
$23.92$
$25$
$25.6$
$15.8$

Average binding energy of deuterium $_{1} H^{2}$ per nucleon $E_{d} = 1.12$ MeV
Average binding energy of $α$ -particle $_{2} H^{4}$ per nucleon $E_{α} = 7.1$ MeV
Energy liberated in the given reaction $Q = 4 (E_{α}) - 2 (E_{d}) - 2 (E_{d})$
$∴$ $Q = 4 (7.1) - 2 (1.12) - 2 (1.12) = 23.92$ MeV