Predicting Vibrational Frequency of D2 from H2

[terp.asessed]

Homework Statement

I am wondering, is it possible to predict the vibrational frequency of D₂ from H₂, supposing vibrational frequency of H₂ is 4400cm^-1?

Homework Equations

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The Attempt at a Solution

From what I know, H₂ has two H's, meaning 2 protons and 2 electrons, whereas D₂ has 2 protons, 2 electrons AND 2 neutrons. So, is the mass the main factor here? If so, how do I 'predict' the vibrational frequency of D₂? Will it be reduced to 1/2 or increased twice that of H₂? Any hint would be appreciated!

 

[Simon Bridge]

Why don't you look up the vibrational modes of hydrogen and duterium molecules and see what actually happens?
Then you will have the basic yes/no answer to the questions you posed above.

Notes:
atomic hydrogen is a bound state of a proton an an electron.
a deuteron is a bound state of a proton and a neutron

A hydrogen molecule is a a pair of covalently bonded hydrogen atoms
A deuterium molecule D₂ is a pair of covalently bonded deuterium atoms.
So you want to know about the molecules.

4400cm^-1 is not a frequency. Frequency has units of inverse-time.

You could start out by modelling each molecules as a mass on each end of a spring.
What happens if you keep the spring constant and length the same, but double the masses?

How could you refine the model to better approximate the atomic systems?
What assumptions are you making?

However - the vibrational modes are usually understood in terms of quantum mechanics.
Where are you at in your education?

 

[terp.asessed]

Wait, come to think of it, 4400cm^-1 is rather a wavelength, NOT frequency (s^-1)! Thank you for correcting me.

Aside, even if the "k" and length are SAME, since both masses have doubled, wouldn't the center of mass still be the same?

Also, you're right--I am currently onto Quantum Mechanics section.

 

[DrClaude]

Wait, come to think of it, 4400cm^-1 is rather a wavelength, NOT frequency (s^-1)! Thank you for correcting me.

Actually, it is the inverse of a wavelength, and is called "wavenumbers". And while it may seem improper to the ears of Simon, it is normal usage among spectroscopists to use cm^-1 to express frequencies or energies, you simply need to multiply by $c$ or $hc$.

Aside, even if the "k" and length are SAME, since both masses have doubled, wouldn't the center of mass still be the same?

You should say "assuming that $k$ and the bond length are the same," as they are the same only to a first-order approximation. The center of mass of two identical atoms will always be midway between them, but the reduced mass will change. Do you know the relation between the reduced mass and the vibrational frequency?

 

[terp.asessed]

Do you know the relation between the reduced mass and the vibrational frequency?

I've learned that reduced mass = μ = m₁m₂/(m₁ + m₂)...so, if the mass of both atoms in a molecule goes UP, then, the reduced mass of the molecule will increase. But, I never thought there was a relationship between reduced mass and vibrational frequency? I thought they were not related?

 

[DrClaude]

I've learned that reduced mass = μ = m₁m₂/(m₁ + m₂)...so, if the mass of both atoms in a molecule goes UP, then, the reduced mass of the molecule will increase. But, I never thought there was a relationship between reduced mass and vibrational frequency? I thought they were not related?

I would be surprised that you were given that exercise without having seen that relation. You should probably review your textbook/notes.

 

[terp.asessed]

Wait, I think I remembered--is it: vibrational frequency = (k/μ)^½/(2πc)?