Does anyone understand how to do the REDUCED MASS OF SYSTEM???

Hi,

Does anyone understand how to do the REDUCED MASS OF SYSTEM and wouldn't mind answering some questions for me???

Basically, I have the following question, but I'm kinda struggling:

1) give an eqn for the reduced mass of the Li2+ system, in terms of mass of nucleus and mass of electron.

As Li in it's atomic state has 3 electrons, and 4 protons+3 nuetrons, and Li2+ has 1 e- and same nuclear contents I've put it as:

u = (Me.7Mn)/(Me+7Mn)

Where Me = mass of electron, and Mn = mass of nucleus

2) Hence calculate the reduced mass and a more accurate energy for n=4 for Li2+

I've calculated the reduced mass to 9.108x10^-31 which I've plugged into R = (e^4.u)/8.Eo^2.h^3.c) but I keep getting the top to be 0... for some reason.

I know the equation, and I know the theoretical method for calculating it, but I'm missing something, and I don't know why

If anyone can help it'd be greatly appreciated.
Thanks
Ourkid

Reply 1

Salain

You need to substitute Rinfinity (109737 or whatever the value for that is) in to it...
R infinity = e^4 Me/8 E^2h^3c

to get the reduced mass you probably multiplied a number like 0.999 by Me, right?
Just take the 0.99 part and multiply that by R infinity, as the rest of the equation is taken in to account by the equation used to get Rinfinity. That way you don't have calculator problems.

Reply 2

Ourkid

Yeah that rings a bell definitely.

The calculation I did was:

u = (Me.7Mn)/(Me+7Mn) = because Li2+ has 1 electron, and 7 protons and nuetrons.

This comes out to be 9.108x10^-31 which must be wrong?

Does anyone know what I've done wrong?

Thank you very much for the reply Salain. :smile: