Answer:
(a) [tex]I=2.33\times 10^{-47}\ kg-m^2[/tex]
(b) [tex]K=1.3\times 10^{-3}\ eV[/tex]
Explanation:
It is given that,
The rotational inertia of a molecule, [tex]I=14000\ upm^2[/tex]
Angular velocity of the molecule, [tex]\omega=4.3\times 10^{12}\ rad/s[/tex]
(a) The molecular weight of a molecule is measured in amu or u.
[tex]1\ amu=1.67\times 10^{-27}\ kg[/tex]
[tex]1 pm=10^{-12}\ m[/tex]
The rotational inertia of the molecule,
[tex]I=14000\times 1.67\times 10^{-27}\times (10^{-12})^2[/tex]
[tex]I=2.33\times 10^{-47}\ kg-m^2[/tex]
(b) The rotational kinetic energy of the molecule is given by :
[tex]K=\dfrac{1}{2}I\omega^2[/tex]
[tex]K=\dfrac{1}{2}\times 2.33\times 10^{-47}\times (4.3\times 10^{12})^2[/tex]
[tex]K=2.15\times 10^{-22}\ J[/tex]
Since, [tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
So, [tex]K=1.3\times 10^{-3}\ eV[/tex]
Hence, this is the required solution.
