find the duration for which an electric bulb of 500W can be kept glowing by the fusion of 100g of deuterium

Solution:Number of atoms in 100 g of deuterium =6.023×1023×1002=3.011×1025Energy released when 2 atoms fuse =3.27 MeVTotal energy released=3.272×3.011×1026×1.6×10-13 J =7.88 ×1013 JEnergy consumed by bulb per second=500Jtime for which the bulb glows =7.88×1013500=0.15753×1013 sec 

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