Question

Two protons and two neutrons combine to form a nucleus of ${}_2{He}^4$. Find the energy released during the process. Take masses of proton, neutron and helium nuclues as $1.007u,1.009u,4.002u$ respectively.

• A. 27.94 MeV
• B. 84.67 MeV
• C. 94.82 MeV
• D. 24.12 MeV

Answer 1

The correct option is A 27.94 MeV

The nuclear reaction is $2_1{p}^1+2_0{n}^1{\to }_2{He}^4$
First, let us find mass defect and then energy.
The mass of $2p$ and $2n$ is $(2\times 1.007+2\times 1.009)=2\times 2.016=4.032u$.
The mass of helium is $=4.002u$.
The mass defect $=4.032-4.002=0.030u$.
$1u$ liberates $931.5MeV$ of energy. The energy equivalent to $0.030u=0.03\times 931.5=27.94MeV$.
The above nuclear reaction is called fusion as lighter nuclei combine together to form a single nuclei.