What is the amount of energy released by deuterium and tritium fusion?

Question

What is the amount of energy released by deuterium and tritium fusion? (A) 60.6 eV (B) 12.6 eV (C) 17.6 eV (D) 28.3 eV

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3288e917a696a7edcf301b4e8957918f-.png" /> The energy released in the process is given by Q=[M21H+M31H-M42He-M10n]c2 =[2.014102 + 3.016050 - 4.002603- 1.008665]uc2 =(0.018884 u)[931.5 (MeV/u)]=17.6 MeV.

Q. What is the amount of energy released by deuterium and tritium fusion?

60.6 eV

12.6 eV

17.6 eV

28.3 eV

The energy released in the process is given by
$Q = [M_{_{_{1}^{2} H}} + M_{_{_{1}^{3} H}} - M_{_{_{2}^{4} He}} - M_{_{_{0}^{1} n}}] c^{2}$
$= [2.014102 + 3.016050 - 4.002603 - 1.008665] u c^{2}$
$= (0.018884 u) [931.5 \frac{M e V}{u}] = 17.6 M e V .$

Q. What is the amount of energy released by deuterium and tritium fusion?

60.6 eV

12.6 eV

17.6 eV

28.3 eV

The energy released in the process is given by Q=[M12​H​​+M13​H​​−M24​He​​−M01​n​​]c2 =[2.014102+3.016050−4.002603−1.008665]uc2 =(0.018884u)[931.5uMeV​]=17.6MeV.

The energy released in the process is given by
$Q = [M_{_{_{1}^{2} H}} + M_{_{_{1}^{3} H}} - M_{_{_{2}^{4} He}} - M_{_{_{0}^{1} n}}] c^{2}$
$= [2.014102 + 3.016050 - 4.002603 - 1.008665] u c^{2}$
$= (0.018884 u) [931.5 \frac{M e V}{u}] = 17.6 M e V .$