Question

Two deuterium nuclei, H, fuse to produce a tritium nucleus, H, and a hydrogen nucleus. A neutral deuterium atom has a mass of 2.014102 u; a neutral tritium atom has a mass of 3.016049 u; a neutral hydrogen atom has a mass of 1.007825 u; a neutron has a mass of 1.008665 u; and a proton has a mass of 1.007276 u. How much energy is released in the process? (1 u = 931.494 MeV/c2)

Answer 1

Answer: The amount of energy released in the process is 4.042 MeV.

Explanation:

The chemical reaction for the fusion of deuterium nucleus follows the equation:

$_1^2\textrm{H}+_1^2\textrm{H}\rightarrow _1^3\textrm{H}+_1^1\textrm{H}$

Atomic mass of the nucleus also contains some mass of the electrons.

Mass of electron in $MeV/c^2$ is $0.511MeV/c^2$

• Calculating the mass of deuterium nucleus:

$m(_1^2\textrm{H})=(2.014102u)\times (931.494MeV/c^2.u)-\text{Mass of electron}$

So, initial mass of the reaction = $2[(1876.124MeV/c^2)-(0.511MeV/c^2)]=3751.226MeV/c^2$

• Calculating the mass of tritium nucleus:

$m(_1^3\textrm{H})=(3.106049u)\times (931.494MeV/c^2.u)-\text{Mass of electron}$

$m(_1^3\textrm{H})=(3.106049u)\times (931.494MeV/c^2.u)-(0.511MeV/c^2)=2803.921MeV/c^2$

So, final mass of the reaction = $2803.921-[(1.007267u\times 931.494MeV/c^2.u)]=3747.184$

Difference between the masses of the nucleus:

$\Delta m=m_{initial}-m_{final}\\\\\Delta m=(3751.226-3747.184)MeV/c^2=4.042MeV/c^2$

Energy released in the process is calculated by using Einstein's equation, which is:

$E=\Delta mc^2$

Putting value of $\Delta m$ in above equation, we get:

$E=(4.042MeV/c^2)\times c^2\\\\E=4.042MeV$

Hence, the amount of energy released in the process is 4.042 MeV.