The isotopic masses of `._(1)H^(2)` and `._(2)He^(4)` are 2.0141 and 4.0026 amu, respectively. Calculate the quantity of energy liberated when two moles of `._(1)H^(2)` undergo fusion to form 1 mol of `._(2)He^(4)`. The velocity of light in vacuum is `3.0 xx 10^(8) m s^(-1)`.

Answer

`2.3xx10^(12)J`

Explanation

`DeltaE=Deltamxxc^(2)` `._(1)H^(2)+``_(1)H^(2)rarr ` `._(2)He^(4)`
Mass defect `=Deltam`
`=` Mass of `2(._(1)H^(2))-` Mass of `He`
`=2xx2.0141-4.0026`
`=4.0282-0026`
`=0.0256g`
`=0.256xx10^(-3)kg`
`c=` speed of light
`=2.998xx10^(8)m s^(-1)`
`DeltaE=Deltamxxc^(20`
`=0.0256xx10^(-3)xx(2.998xx10^(8))^(2)`
`=0.23xx10^(13)J=2.3xx10^(12)J`