Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 10

Q. 10.1

Estimate the value of E_{ rot } for the lowest rotational energy state of N _{2}, which has a bond length 0.110 nm.

Strategy If we consider the nitrogen atoms to be point masses (each with mass m) separated by a distance R (see Figure 10.4), then the rotational inertia about an axis passing through the center of the molecule and perpendicular to the line joining the atoms is

I=m\left(\frac{R}{2}\right)^{2}+m\left(\frac{R}{2}\right)^{2}=\frac{m R^{2}}{2}

With this numerical estimate of rotational inertia I, the energy levels follow from Equation (10.2).

E_{ rot }=\frac{\hbar^{2} \ell(\ell+1)}{2 I} (10.2)

Step-by-Step

Verified Solution

For nitrogen, m=2.33 \times 10^{-26} kg \text {, and } R=1.10 \times 10^{-10} m.

Thus,

 

\begin{aligned}I &=\frac{\left(2.33 \times 10^{-26} kg \right)\left(1.10 \times 10^{-10} m \right)^{2}}{2} \\&=1.41 \times 10^{-46} kg \cdot m ^{2}\end{aligned}

 

For \ell=1,

 

\begin{aligned}E_{ rot } &=\frac{2 \hbar^{2}}{2 I}=\frac{\hbar^{2}}{I}=\frac{\left(1.055 \times 10^{-34} J \cdot s \right)^{2}}{1.41 \times 10^{-46} kg \cdot m ^{2}} \\&=7.89 \times 10^{-23} J =4.93 \times 10^{-4} eV\end{aligned}

 

Remember, however, that this is just the energy of the lowest state. The energy of the \ell=4 \text { state }[\text { with } \ell(\ell+1)=20] is 10 times higher, or about 0.005 eV. This energy is still at least two orders of magnitude less than the energy associated with an electronic transition in hydrogen (the kind that produces visible photons). Therefore we can expect photons generated by transitions between adjacent rotational states to be in the infrared or microwave portion of the spectrum.