13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as


H12  + H12  H e 2 3 + n+3.27 MeV

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7 months ago

13.19 The given fusion reaction is

 H12+ H12  H e 2 3 + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023*1023
 atoms

Hence 2 kg of deuterium contains = 
6.023*10232 *2*103atoms = 6.023 *1026 atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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