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' .