1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hydrogen vs. Deutron

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    I am wondering, is it possible to predict the vibrational frequency of D2 from H2, supposing vibrational frequency of H2 is 4400cm-1?

    2. Relevant equations

    3. 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!
  2. jcsd
  3. Nov 12, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    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.

    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?
  4. Nov 12, 2014 #3
    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.
  5. Nov 12, 2014 #4


    User Avatar

    Staff: Mentor

    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##.

    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?
  6. Nov 12, 2014 #5
    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?
  7. Nov 12, 2014 #6


    User Avatar

    Staff: Mentor

    I would be surprised that you were given that exercise without having seen that relation. You should probably review your textbook/notes.
  8. Nov 12, 2014 #7
    Wait, I think I remembered--is it: vibrational frequency = (k/μ)½/(2πc)?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted