Calculating Reduced Mass of NO+ Molecular Ion: Missing Electron Considerations

[bcjochim07]

Homework Statement

I want to calculate the reduced mass of a molecular ion, NO^{+} (singly-ionized nitrogen monoxide molecule) in atomic units. My question is, how would I take into account the missing electron in this calculation.

Homework Equations

mass of N in a.u. is about 25525 a.u.

mass of O in a.u. is about 29157 a.u.

mass of electron in a.u. is 1 a.u.

The Attempt at a Solution

--Would it be reasonable to arbitrarily say that the electron was taken from the O atom and thus the reduced mass is given by the following:

(29156)(25525)/(29156+25525) = 13610.0 a.u.

--If I say that the electron was taken from N, I would get the following:

(29157)(25524)/(29157+25524) = 13609.9 a.u. --- not much different

Or I could find the reduced mass of NO and then subtract one:

(29157)(25525)/(29157+25525) - 1 = 13609.2 a.u. --- again not too different


Any ideas on which of these is the most correct way to do the calculation, or if there is another better way to approach it? Thanks.

 

[kuruman]

Or you could ignore the electron altogether and get 13610.2 for the reduced mass. So how accurate do you want your answer to be? When you say "the mass of N is about 25525 a.u.", what does the word "about" mean? If the mass of N is 25525 ± 1 a.u. then you can safely ignore the mass of the electron and stop agonizing about what to do with it.

 

[Jack the Stri]

May I just ask where you get your calculations from? I don't see the logic in them and calculating the mass of neutral NO doesn't seem to compute from it... (plus, your reduced mass is less than the mass of the individual atoms)

 

[kuruman]

May I just ask where you get your calculations from? I don't see the logic in them and calculating the mass of neutral NO doesn't seem to compute from it... (plus, your reduced mass is less than the mass of the individual atoms)

Does this surprise you? Reduced mass is given by

$$\mu=\frac{m_1*m_2}{m_1+m_2}$$

When the masses are equal, and they are nearly equal here, the reduced mass is half of either mass. I believe that's why it is called "reduced".

 

[Jack the Stri]

Whoops, my bad :-)