1.

Calculate the energy generated in kWh, when 100g of `._(3)Li(7)`. are converted into `._(2)He(4)` by proton bombardment. Given mass of `._(3)Li(7)=7.0183a.m.u ,` mass of `._(2)He(4)=4.0040a.m.u`, mass of `._(1)H(1)atom=1.0081a.m.u.`

Answer» Correct Answer - `6.5475xx10^(6)kWh`
Here, E=?
The nuclear reaction is
`._(3)Li^(7)+._(1)H^(1)to 2 ._(2)He^(4)+Q`
Mass defect, `Deltam=7.0183+1.0081-2xx4.0040`
`=0.0184 a.m.u.`
`Q=0.00184xx931MeV=17.13MeV`
Number of nuclei in 100g of `._(3)Li^(7)`
`=(6.023xx10^(23))/7xx100`
`=8.6xx10^(24)`
Total energy released when 100 g of `._(3)Li^(7)` is converted into Helium
`=17.13xx8.6xx10^(24)MeV`
`=17.13xx8.6xx1.6xx10^(-13)xx10^(24)J`
As `1kWh=3.6xx10^(6)J`
`:. E=(17.13xx8.6xx1.6xx10^(11))/(3.6xx10^(6))kWh`
`E=6.5475xx10^(6)kWh`


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