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Examine table 3 and express (a) energy required to break on bond in DNA in eV, (b) typical energy of a proton in a nucleus in eV, (c ) energy released in burning 1000 kg of coal in kilocalories. |
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Answer» Solution :(a) Energy required to BREAK one bond of DNA `=(10^(-20)J)/(1.6xx10^(-19)Je//V)=0.06eV` (B) Typical energy of a proton in a nucleus `=(10^(-13)J)/(1.6xx10^(-19)J//eV)=625000eV` (c ) Energy released in buming 1000 kg of coal `=(3xx10^(10)J)/(4.2xx10^(3)"J/kcal")=7142857.14" kcal"` |
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