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### Topic: Microwave spectroscopy  (Read 7788 times)

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#### fogy

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• Mole Snacks: +0/-0 ##### Microwave spectroscopy
« on: December 31, 2011, 09:41:12 AM »
The rotational constant B for H35Cl was measured to be 10.5909 cm-1. Calculate the values for H37Cl and 2D35Cl.

B=h/(8pi^2)(I)(c)

10.5909= h/(8pi^2)(I)(c)

do I find the moment of inetira (I) this way for H35CL? ...and then calculate the reduced mass for H37CL? Then giving me a new value for I by the equation ..r squared = u/I

Where or how would i get the value for the bond length/ internecleur distance (these are the same right?) so i could calculate the new I

Maybe im doing this whole question wrong.

Help would be appriciated
Thanks Thanks « Last Edit: December 31, 2011, 10:01:58 AM by fogy »

#### Borek ##### Re: Microwave spectroscopy
« Reply #1 on: December 31, 2011, 11:51:24 AM »
Assume bond length doesn't change. You don't need to know its exact value.
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#### fogy

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• Mole Snacks: +0/-0 ##### Re: Microwave spectroscopy
« Reply #2 on: January 02, 2012, 07:30:12 AM »
Assume bond length doesn't change. You don't need to know its exact value.

H35Cl: B= 10.5909cm^-1

reduced mass for H37CL is calculated to be : 2.8003*10^-54Kg

Now im thinking that if i want to calculate B for this specific compound il need to use B= h/8pi^2 (I)(c) Missing I here so need to calculate that by using : I=(R^2)(U)
But then see that i need R ( bond length figure) to be able to calculate I and so B.

There is obviously another way of doing this so that you dont need the bond length figure?

#### Borek ##### Re: Microwave spectroscopy
« Reply #3 on: January 02, 2012, 08:10:05 AM »
Starting with

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

and remembering that

$$I = \mu R^2$$

you can easily get

$$\frac {B_{H^{35}Cl}} {B_{H^{37}Cl}} = \frac {\mu_{H^{37}Cl}} {\mu_{H^{35}Cl}}$$

(edit: that is, assuming R is constant).
« Last Edit: January 02, 2012, 10:05:34 AM by Borek »
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#### fogy

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• Mole Snacks: +0/-0 ##### Re: Microwave spectroscopy
« Reply #4 on: January 02, 2012, 12:12:41 PM »
I understand now, thanks a mil!! 