What Are the Wavenumbers for the First Three Rotational Transitions of CO?

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SUMMARY

The discussion focuses on calculating the wavenumbers for the first three rotational transitions of carbon monoxide (CO) using quantum mechanics principles. The bond length of CO is given as 1.128 Å, and the relevant equations include the moment of inertia (μ), rotational constant (B), and energy transitions (ΔE). The user successfully calculated the moment of inertia and the rotational constant, arriving at a value of 1.93 cm-1 for the first transition. Further assistance is needed to apply the ΔE equation to find the wavenumbers for the subsequent transitions.

PREREQUISITES
  • Understanding of quantum mechanics principles related to molecular rotation
  • Familiarity with the equations for moment of inertia and rotational constants
  • Knowledge of Planck's constant and its application in energy calculations
  • Basic skills in unit conversion, particularly between Ångströms and meters
NEXT STEPS
  • Learn how to apply the ΔE equation to calculate rotational energy levels for diatomic molecules
  • Research the concept of rotational transitions in quantum mechanics
  • Study the relationship between wavenumbers and energy transitions in spectroscopy
  • Explore advanced topics in molecular spectroscopy, focusing on CO and similar diatomic molecules
USEFUL FOR

This discussion is beneficial for students studying quantum mechanics, particularly those focusing on molecular spectroscopy and rotational transitions. It is also useful for educators and researchers interested in the rotational dynamics of diatomic molecules like carbon monoxide.

Klestine
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Homework Statement


The bond length of CO is 1.128 Å. At what wavenumbers do the first three rotational transitions appear


Homework Equations


μ= (m1m2/m1+m2)
I = μr0^2
B= h/(8∏^2)IbC
ΔE = 2B (J+1)


The Attempt at a Solution



For the first equation I got

(12 x 16) / 28 = [(6.857 /1000)/ avagadro number] = 1.138 x 10^-26

For the second equation I got

(1.138 x 10^-26) x (1.128 x 10^-10)^2 = 1.44 x 10^-46 kg/m^2

For the third equation I got

planck's constant / [(8∏^2) x (1.44 x 10^-46 ) x (speed of light)] = 1.93cm^-1



I honestly don't know what to do after this please help
really appreciated and thank you very much
 
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You're doing fine. What about the fourth equation you listed?
 
DrClaude said:
You're doing fine. What about the fourth equation you listed?

Thats the problem, I know the answer is in that equation but I just do not know how to use it to get the values for the first 3 rotational energy levels. Just don't know to tackle that equation.Thanx for the reply
 

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