Question

The binding energy per nucleon of deuteron is $1.2MeV$ and that of the helium atom is $7.1MeV$. What is the energy released, if two deuteron atoms combine to form a single helium atom?

• A. 17.8MeV
• B. 19.5MeV
• C. 21,2MeV
• D. 23.6MeV

Answer 1

The correct option is D 23.6MeV

$\begin{array}{c}\text{Given -}\ast BE\text{per nucleon of deuteron}=1.2MeV\\ \text{and of Helium}=7.1MeV\\ {}_1^{2}H+{}_1^{2}H{⟶}_4^{2}He\end{array}$

$\begin{array}{c}\Rightarrow \text{Energy released = BE of Helium -2.BE of Deutron.}\\ \text{BE}=BE\text{ber nucleons}X\text{Num of nucleons.}\\ \begin{array}{cc}\Rightarrow E& =[4\times 7.1]-2\times [1.2\times 2]\text{Mev}\\ E& =23.6MeV\end{array}\end{array}$