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Calculate and compare the energy released by :(a) Fusion of 1 kg of hydrogen deep within the sun, and (b) The fission of 1 kg of 92U235 in a fission reactor. |
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Answer» In sun, Four hydrogen nuclei fuse to form a helium nucleus with the release of 26 MeV energy. ∴ Energy released by fusion of 1kg of hydrogen : E1 = \(\frac{6\times 10^{23}\times 10^3}{4}\) × 26 MeV. E1 = 39× 1026Mev As, Energy released in fission of one atom of 92U235 = 200 MeV. ∴ Energy released in fission of 1kg of 92U235 : E2= \(\frac{6\times 10^{23}\times 10^3}{235}\) × 26 MeV. E2 = 5.1× 1026MeV ∴ \(\frac{E_1}{E_2}\) = \(\frac{39\times 10^{26}}{5.1\times 10^{-26}}\) = 7.65 i.e. energy released in fusion is 7.65 times the energy released in fission. |
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