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

AI Thread Summary
The discussion focuses on calculating the bond length of carbon monoxide (CO) from the J=0 to J=1 transition frequency of 1.153x10^5 MHz. The initial calculation yielded a bond length of 2520 pm, significantly larger than the observed value of 113 pm. Users suggested checking the formulas for rotational energy and the moment of inertia, indicating that the correct approach involves using the relationship between energy and frequency to derive bond length. The correct bond length calculation should yield approximately 115.3 pm after proper manipulation of the equations. Attention to unit conversion and calculation accuracy is emphasized as critical for obtaining the correct result.
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Homework Statement



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.

Homework Equations



2B = h/(4π2I)

μ = m1*m2/(m1+m2)

I = μr2

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!
 
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cep said:
2B = h/(4π2I)

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

\bar B = \frac {h} {8\pi^2cI}

or

B = \frac {\hbar} {2I}

Neither fits what you wrote.
 
I've tried both of those-- still orders of magnitude too large.
 
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!
 
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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