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Calculate CO bond length from J=0 to J=1 transition

  1. Oct 26, 2011 #1

    cep

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    1. The problem statement, all variables and given/known data

    The J=0 to J=1 transition fro 12C16O carbon monoxide occurs at 1.153x105 MHz. Calculate the value of the bond length in CO.

    2. Relevant equations

    2B = h/(4π2I)

    μ = m1*m2/(m1+m2)

    I = μr2

    3. The attempt at a solution

    I = 6.61x10-34/ (4*π2*1.153x108) = 1.45x10-43 kg m2

    μ = 12.01*16.00/(12.01+16.00) *1.661x10-27 = 1.14x10-26kg

    r = √(I/μ) = √(1.45x10-43/1.14x10-26) = 2.52x10-9 m or 2520 pm, which is significantly larger than the observed value of 113 pm.

    Does anyone see where I went wrong?

    Thanks!
     
  2. jcsd
  3. Oct 26, 2011 #2

    Borek

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    Staff: Mentor

    My QM is so rusty I am trying to not touch it as it may fell apart, but I just checked in wikipedia and it is either

    [tex]\bar B = \frac {h} {8\pi^2cI}[/tex]

    or

    [tex]B = \frac {\hbar} {2I}[/tex]

    Neither fits what you wrote.
     
  4. Oct 26, 2011 #3

    cep

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    I've tried both of those-- still orders of magnitude too large.
     
  5. Mar 19, 2012 #4
    Even though this is an old question I figured I'd post an answer for people who might have the same question and end up here.

    The rotational energy for a specific level J, EJ = J(J+1)[((m1 + m2)*h2)/(8m1m2R2) = J(J+1)B

    In this case B = ((m1 + m2)*h2)/(8m1m2R2)

    The transition from J = 0 → J = 1 then would be [1(1+1)B] - [0(0+1)B] = 2B

    1.153x105 MHz is 1.153x1011 Hz. Solving for E from E = hv where h is still planck's constant and v is the frequency gives 7.64x10-23 Joules which is equal to 2B as shown above.

    7.64x10-23 Joules = 2B = 2[((m1 + m2)*h2)/(8m1m2R2)]

    From here you just manipulate the equation to solve for R then plug in the values for m1, m2, E, and h. The value for R (not a radius mind you, but a bond length) should be ~115.3 pm.

    Oh and if you still aren't getting the right answer, make sure the check your units!
     
  6. Nov 29, 2012 #5
    You were right only missed a factor of 3 by 1.153x10^11, the rest check your calculator. I know its been a while :cry:
     
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