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)

$Q=\displaystyle \sum B.E. \, of \, products- \, \displaystyle \sum B.E. \, of \, reactants$
The equation for two deuterons combining to form Helium nucleus is given as $H^{2}+_{1}H^{2} \rightarrow _{2}H^{4} \, $
$Energy \, released=Q=4\left[B . E . \left(_{2} H^{4}\right)\right] \, -4\left[B . E . \left(_{1} H^{2}\right)\right]$
$=4\times 7-4\times \text{1.1}$
$=28-\text{4.4}$
The energy released in this process is $=\text{23.6}MeV$