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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.` |
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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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