Question

A molecule has a rotational inertis of 14,000 u.pm^2 and

Answer 1

Answer:

(a) $I=2.33\times 10^{-47}\ kg-m^2$      

(b) $K=1.3\times 10^{-3}\ eV$

Explanation:

It is given that,

The rotational inertia of a molecule, $I=14000\ upm^2$

Angular velocity of the molecule, $\omega=4.3\times 10^{12}\ rad/s$

(a) The molecular weight of a molecule is measured in amu or u.

$1\ amu=1.67\times 10^{-27}\ kg$

$1 pm=10^{-12}\ m$

The rotational inertia of the molecule,

$I=14000\times 1.67\times 10^{-27}\times (10^{-12})^2$

$I=2.33\times 10^{-47}\ kg-m^2$      

(b) The rotational kinetic energy of the molecule is given by :

$K=\dfrac{1}{2}I\omega^2$

$K=\dfrac{1}{2}\times 2.33\times 10^{-47}\times (4.3\times 10^{12})^2$    

$K=2.15\times 10^{-22}\ J$

Since, $1\ eV=1.6\times 10^{-19}\ J$

So, $K=1.3\times 10^{-3}\ eV$

Hence, this is the required solution.