Question

The binding energies per nucleon for deuterium and helium are 1.1 MeV and 7.0 MeV respectively. What energy in joules will be liberated when `10^(6)` deuterons take part in the reaction.

Answer 1

`[(B.E)/(A)]_(D)=1.1 MeV , [(B.E)/(A)]_(He)=7.0 MeV`
For deuterium `(._(1)^(2)H)`,
A = 2
For He `(._(2)^(4)He), A = 4`
`[(B.E)/(2)]_(D)=1.1 MeV rArr [B.E]_(D)`
`= 2xx1.1 MeV = 2.2 MeV`
`[(B.E)/(4)]_(He)=7.0 MeV rArr [B.E]_(He)`
`= 4xx7.0 MeV = 28 MeV`
We know `._(1)H^(2)+ ._(1)H^(2) rarr ._(2)He^(4)`
Energy released = B.E of `10^(6)` deuterons - B.E of `(1)/(2)xx10^(6)` Helium atons
`B.E = 2.2xx10^(6)xx(1)/(2)xx10^(6)xx28`
`= 10^(6)(2.2-14)`
`= -11.8xx10^(6)MeV`
`=-11.8xx10^(6)xx1.6xx10^(-13)J`
`=- 18.88xx10^(-7)J`
(- ve sign indicates that energy is released)
`therefore` Energy released `= 18.88xx10^(-7)J`