Mass Spectrometer and Electric Field Problem

[jena]

Hi,

My Question:

Suppose the electic field between the two electric plates in the mass spectrometer is 2.48 x 10^4 V/m and the magnetic fields B=B'=0.68 T. The source contains carbon isotopes of mass number 12,13,14 from a long dead piece of tree(To estimate atomic masses multiply by 1.67 x 10^.27 kg). How far apart are the lines formed by the singlely charged ions of each type on the photographic film? What if the ions were doubly charged?

My work:

For carbon isotope 12

m=(qBB'r)/E

12(1.67 x 10^.27 kg)=((1.6 x 10^.19 coul)(0.68 T)(0.68T)(r))/(2.48 x 10^4 V/m)

r=.0067m

Is this correct

As for the the ions being doubly charged wouldn't I just double, q.

Thank You

 

[Astronuc]

Your answer of r = 0.0067 m is correct for C¹², but remember that if a particle travels a semicircle, the separation between lines will be the difference in the diameter (or 2*r) of the ion trajectories.

If the atoms are doubly ionized, then q is doubled. The electric force is twice, but so is the magnetic field.

Remember, if B is the field in the electrostatic field, and the forces of the electric field and magnetic field balance, then

q E = q v B, or v = E/B, thus v is independent of charge.

B' is the deflection field and r = mv / qB' .

 

[quicknote]

Your calculations for the first part seem to be correct. To answer your second question, yes, if the ions were doubly charged, you would simply double the charge (q) in your calculations. This would result in the ions being closer together on the photographic film, as they would experience a stronger force in the electric field. However, the exact distance between the lines formed would depend on the specific charge of each isotope (q/m ratio), which may differ for different isotopes. It would also depend on the strength of the magnetic field, as this affects the radius of the ion's path.