[CHM 331] 9.26

[Andrew J. Pounds]

On 09/15/2012 12:09 AM, Christine.O.Conroy wrote:
> For 9.26 a,
>
>
> In class we found that
> k=4 pi^2 v^2 m
>
> In the book it says
> m(mu)=m1m2/(m1+m2)
>
> So for 35 Cl2 I found that
>
> k=4 (pi^2) (5.6)^ 2 m
>
> for m I found the mass of one Cl atom by dividing .035 by avogandros 
> number getting 5.812E-26 and then found mu to be (5.812E-26)^2 / 
> (2*5.812E-26) and got a value of 2.906E-26 kg.
>
> so k =4 pi^2 (5.6)^2 (2.906E-26)
>
>
> But that is not the answer in the back of the book.
>
> I looked online and it seems that some equations include c or the 
> speed of light in the equation (ie k= m(2*pi*v*c)^2 but I did't see 
> this in the notes or in the book and regardless adding it does not 
> give me the correct answer.
>
>
> Could you help?

The first problem I see is your conversion to reciprocal meters.  
Because it is an inverse relationship you multiply to 100 (to get 56000 
), not divide by 100.



-- 
Andrew J. Pounds, Ph.D.  (pounds at theochem.mercer.edu)
Associate Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627

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