__Question 7.20__:^{–26}kg and a moment of inertia of 1.94×10

^{–46}kg m

^{2}about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

__Solution__:Mass of an oxygen molecule,

*m*= 5.30 × 10

^{–26}kg

*I*= 1.94 × 10

^{–46}kg m

^{2}

Velocity of the oxygen molecule,

*v*= 500 m/s

The separation between the two atoms of the oxygen molecule = 2

*r*

Mass of each oxygen atom = m/2

Hence, moment of inertia

*I*, is calculated as:

(m/2)r

^{2}+ (m/2)r

^{2}= mr

^{2}

r = ( I / m)

^{1/2}

(1.94 X 10

^{-46}/ 5.36 X 10

^{-26})

^{1/2}= 0.60 X 10

^{-10}m

It is given that:

KE

_{rot}= (2/3)KE

_{trans}

(1/2) I ω

^{2}= (2/3) X (1/2) X mv

^{2}

mr

^{2}ω

^{2}= (2/3)mv

^{2}

ω = (2/3)

^{1/2}(v/r)

= (2/3)

^{1/2}(500 / 0.6 X 10

^{-10}) = 6.80 X 10

^{12}rad/s

The given solution is wrong.check the line where you are calculating the moment of inertia.

ReplyDeleteYou should write mr^2 as I. To avoid confusion.

DeleteYou should write mr^2 as I. To avoid confusion.

Delete