Calculating Wavelength of Emitted Radiation in Carbon Monoxide Molecule

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SUMMARY

The discussion focuses on calculating the wavelength of emitted radiation when a carbon monoxide (CO) molecule transitions from the first excited rotational state (l=1) to the ground state (l=0). The moment of inertia (I) for the CO molecule is given as 2.23 x 10^-46 kgm². The energy difference between these states is calculated using the formula E = l(l+1)h_bar² / 2I, leading to the correct wavelength of emitted radiation being 2.6 mm. The calculations involve fundamental constants such as Planck's constant and the speed of light.

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Homework Statement



The molecular bond between the carbon and oxygen atoms in the diatomic carbon monoxide molecule can be treated as a spring with elastic constant k. The carbon atom has a mass of 12 amu and the oxygen atom has a mass of 16 amu, and the average separation of the two nuclei in the carbon monoxide molecule is 1.4x10^-10m.

A carbon monoxide molecule makes a transition from the first excited rotational state to the ground state. What is the wavelength of the emitted radiation?


Homework Equations



E = l(l+1)h_bar^2 / 2I

E = energy
l = quantum number
h_bar = Plancks constant / 2*pi
I = moment of inertia

E = hc/lambda

h = placks constant
c = speed of light in vacuum
lambda = wavelength

The Attempt at a Solution



I calculated the moment of inertia I in a previous part of the question and it is correct.

I = 2.23 x 10^-46 kgm^2

I assumed that ground state means that l=0 and first excited state means that l=1. I calculated the energy difference between these two states. Then calculated the debroglie wavelength. But I didn't get the correct answer.

Correct answer should be : 2.6 mm
 
Physics news on Phys.org
yes, l=0 is the ground state & l=1 is the excited state.
 

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