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5.Work, Energy, Power and Collision
medium
Express
$(a)$ The energy required to break one bond in $DNA$ in $eV$
$(b)$ The kinetic energy of an air molecule $\left(10^{-21} \;J \right)$ In $eV$
$(c)$ The daily intake of a human adult in kilocalories.
Option A
Option B
Option C
Option D
Solution
$(a)$ Energy required to break one bond of $DNA$ is
$\frac{10^{-20} J }{1.6 \times 10^{-19} J / eV } \simeq 0.06 eV$
Note $0.1 eV =100 meV$ ( $100$ millielectron volt).
$(b)$ The kinetic energy of an air molecule is
$\frac{10^{-21} J }{1.6 \times 10^{-19} J / eV } \simeq 0.0062 eV$
This is the same as $6.2 meV$
$(c)$ The average human consumption in a day is
$\frac{10^{7} J }{4.2 \times 10^{3} J / kcal } \simeq 2400 kcal$
Standard 11
Physics
