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Examine table 3 and express (a) energy required to break one bond in DNA in eV, (b) typical energy of a proton in a nucleus in eV, (c ) energy released in burining 1000 kg of coal in kilicalories |
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Answer» Solution :(a)Energy REQUIRED to break one bond of DNA `=(10^(-20))/(1.6xx10^(-19)J//eV)=0.06eV` (B) Typical energy of a PROTON in a nucleus `=(10^(-13))/(1.6xx10^(-19)J//eV)=625000eV` (c ) Energy RELEASED in burning 1000 kg of coal `=(3xx10^(-10))/(4.2xx10^(3)J//Kcal)=7142857. 14 kcal` |
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