Examine Tables 61. -6.3 and express (a) the energy required to break one bond in DNA in eV, (b) The kinetic energy of an air molecule `(10^(-21)J)` in eV , (c ) The daily intake of a human adult in kilocalories .

Explanation

(a) Energy required to break one bond of DNA is `E = ((10^(-20)J))/(1.6 xx10^(-19)J//eV) = 0.0625 eV `
(b) The kinetic energy of an air molecule is `1.6xx10^(19)J = 1 eV `
` 10^(-21)J = ?`
` :. K = (10^(-21))/(1.6xx10^(-19))`
` :. K = 0.00625 eV `
(c ) The average human consumption in day is `4.2 xx 10^(3) J = 1 kcal `
` 10^(-7) J =? `
`H = (10^(7))/(4.2 xx10^(3)) = 2400 ` K cal
` :. H = 0.23809 xx10^(4)` ,
` :. H approx k cal = 2400 ` k cal