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Distance between the atoms in HCl

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data
    At transition from first excited state of rotator to ground state the molecule of HCl shoots out a photon with ##4,8\cdot 10^{-4}m##. How far apart are the atoms in the molecule?


    2. Relevant equations



    3. The attempt at a solution

    What is wrong with:

    ##\Delta E=E_\gamma ##

    ##\frac{\hbar ^2}{2m_rr^2}|l_f(l_f+1)-l_i(l_i+1)|=\frac{hc}{\lambda }##

    Where ##f## indicates final state and ##i## for initial state and ##m_r=\frac{1\cdot 35.45}{1+35.45}=0,97au##.

    ##l_f=0## and ##l_i=1##, which gives me

    ##\frac{\hbar ^2}{2m_rr^2}(2)=\frac{hc}{\lambda }##

    and finally

    ##r=\sqrt{\frac{hcm_rc^2}{\lambda \hbar ^2c^2}}=7.64 nm## instead of ##0.13 nm##.

    What am I doing wrong here?
     
  2. jcsd
  3. Apr 13, 2014 #2
    Your reasoning is correct, however your expression for [itex]r[/itex] seems to be off, I get:
    [itex]r=\sqrt{\frac{h\lambda}{4π^{2}mc}}[/itex]
    which yields the desired answer.
     
  4. Apr 14, 2014 #3
    Ah, of course... My result is actually ##r^{-1}##. Thanks for that!
     
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