Question

A nitrogen molecule has two nitrogen atoms separated by a distance of $1.3\times {10}^{-10}m$. Find its moment of inertia about the centre of mass. If its rational kinetic energy is $4\times {10}^{-21}J$. Then compute its frequency of rotation about the centre of mass. the mass of a nitrogen atom is $14$ amu ($1\text{amu}=1.66\times {10}^{-27}kg$)

• A. (i) $19.6\times {10}^{-47}kg{m}^2$; (ii) $1.0\times {10}^{12}Hz$
• B. (i) $1.96\times {10}^{-49}kg{m}^2$; (ii) $1.0\times {10}^{13}Hz$
• C. (i) $1.96\times {10}^{-40}kg{m}^2$; (ii) $1.0\times {10}^{12}Hz$
• D. (i) $1.34\times {10}^{-46}kg{m}^2$; (ii) $1.0\times {10}^{12}Hz$

Answer 1

The correct option is A (i) $19.6\times {10}^{-47}kg{m}^2$; (ii) $1.0\times {10}^{12}Hz$

The mass of one nitrogen atom is$=14\times \frac{1}{6.023\times {10}^{23}}\times {10}^{-3}\Rightarrow 23.24\times {10}^{-27}kg$

MoI of nitrogen molecule,$I=2\times 23.24\times {10}^{-27}\times {\left(\frac{1.3\times {10}^{-10}}{2}\right)}^2=19.6378\times {10}^{-47}kg-m^2$

Now,

$E_k=\frac{1}{2}I{\omega }^2$

$\Rightarrow 4\times {10}^{-21}=\frac{1}{2}\times 19.6378\times {10}^{-47}\times {\omega }^2$

$\Rightarrow \omega =0.638\times {10}^{13}$

Now,

$f=\frac{\omega }{2\pi }=\frac{0.638\times {10}^{13}}{2\times 3.14}={10}^{12}Hz$