Question

Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron $=\text{1.1}MeV$ and for helium $=\text{7.0}MeV$ )

• A. $\text{23.6}MeV$
• B. $\text{30.2}MeV$
• C. $\text{25.8}MeV$
• D. $\text{32.4}MeV$

Answer 1

Answer: (A)

$Q=\sum B.E.ofproducts-\sum B.E.ofreactants$
The equation for two deuterons combining to form Helium nucleus is given as $H^2{+}_1{H}^2{\to }_2{H}^4$
$Energyreleased=Q=4\left[B.E.\left({}_2{H}^4\right)\right]-4\left[B.E.\left({}_1{H}^2\right)\right]$
$=4\times 7-4\times \text{1.1}$
$=28-\text{4.4}$
The energy released in this process is $=\text{23.6}MeV$