Question

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

isspinning at an angular speed of 4.3 X 10^12 rad/s.
a.) Express the rotational inertia in kg.m^2.
b.) Calculate the rotational kinetic energy
inelectron-volts.

Answer 1

(a)
`I=2.33* 10^(-47)\ kg-m^2`

(b)
`K=1.3* 10^(-3)\ eV`

Step-by-step explanation:

It is given that,

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

Angular velocity of the molecule,
`\omega=4.3* 10^(12)\ rad/s`

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

`1\ amu=1.67* 10^(-27)\ kg`

`1 pm=10^(-12)\ m`

The rotational inertia of the molecule,

`I=14000* 1.67* 10^(-27)* (10^(-12))^2`

`I=2.33* 10^(-47)\ kg-m^2`

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

`K=(1)/(2)I\omega^2`

`K=(1)/(2)* 2.33* 10^(-47)* (4.3* 10^(12))^2`

`K=2.15* 10^(-22)\ J`

Since,
`1\ eV=1.6* 10^(-19)\ J`

So,
`K=1.3* 10^(-3)\ eV`

Hence, this is the required solution.