Question

The binding energy/nucleon of deuteron ${(}_1{H}^2)$ and the helium atom ${(}_{2}H{e}^4)$ are $1.1$ MeV and $7$MeV respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is.

• A. $26.9$MeV
• B. $25.8$MeV
• C. $23.6$MeV
• D. $12.9$MeV

Answer 1

Answer: (C)

$23.6$MeV

The nuclear reaction is: ${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4$

Binding energy of deutron, $E_d=1.1\times 2=2.2$ MeV

Binding energy of helium, $E_{He}=7\times 4=28$ MeV

Energy released, $Q=E_{He}-2E_d$

$\therefore$ $Q=28-2\left(2.2\right)=23.6$ MeV