Question

The oxygen molecule has a mass of $5.30 \times 10^{-26}\, kg$ and a moment of inertia of $1.94 \times 10^{-46}\, kg - m ^{-2}$ 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 is $KE$ of rotation is $\frac{2}{3}$ of its KE translation. Find the average angular velocity of the molecule.

• A. $3.75 \times 10^{12}\, rad / s$
• B. $5.75 \times 10^{12}\, rad / s$
• C. $9.75 \times 10^{12}\, rad / s$
• D. $6.75 \times 10^{12}\, rad / s$

Answer 1

Answer: (D)

Mass of oxygen molecule
$(M)=5.30 \times 10^{-26} kg$
Moment of inertia $I=1.94 \times 10^{-46} kg - m ^{2}$
Mean speed of the molecule $(v)=500\, m / s$
Given, $KE$ of rotation $=\frac{2}{3} \times KE$ of translation
$\frac{1}{2} I\omega^{2} =\frac{2}{3} \times \frac{1}{2} M v^{2}$
or $\omega =\sqrt{\frac{2 M v^{2}}{l}}$
$=\sqrt{\frac{2 \times 5.30 \times 10^{-26} \times(500)^{2}}{1.94 \times 10^{-46}}}$
$=1.35 \times 10^{10} \times 500$
$=6.75 \times 10^{12}\, rad / s$