Question

Energy released in the nuclear fusion reaction is :
${}_1{\text{H}}^2{+}_1{\text{H}}^3\to {}_2{\text{He}}^4+{}_0{\text{n}}^1$
Atomic mass of some species are given below.
${}_1{\text{H}}^2=2.014,{}_1{\text{H}}^3=3.016$
${}_2{\text{He}}^4=4.003,{}_0{\text{n}}^1=1.009\text{amu}$.

• A. $8.30\text{eV}$
• B. $16.758\text{MeV}$
• C. $500\text{J}$
• D. $4\times {10}^6\text{kcal}$

Answer 1

The correct option is B $16.758\text{MeV}$

Mass defect = $(2.014+3.016)-(4.003+1.009)=0.018\text{a.m.u}$
Now, $1\text{a.m.u.}\approx 931\text{MeV}$
Energy released = $0.018\times 931=16.758\text{MeV}$.