Tough College physics problem...someone please lead me in the right direction...?

Question

A singly ionized 24Mg ion (mass: 3.983x10^kg) is accelerated through a 2.5kV potential difference & deflection in a magnetic field of 0.0557 T in a mass spectrometer.(A) Find the radius of curvature of the orbit for the ion. (B) What is the difference in radius for 26Mg ions and for 24Mg ions? (assume that their mass ratio is 26:24) OK SOMEONE HELP... and by chance does anyone know a good website to help with college physics?

Answer 1

Figure out how much electrical potential energy it has at the beginning. This energy is transformed into kinetic energy after traveling through the electric field. You can determine the speed from the kinetic energy. You can then use the speed to figure out the radius of curvature when It is in the magnetic field.

Answer 2

Apply conservation of energy, assume ion starts at rest
Ei = Ef
(charge) (Potential Diff) = Kinetic energy
since KE = 1/2 mv^2 you can solve for v

The radius of curvature is given by
r = (momentum)/(charge x Magnetic field)

for part b, we note that p larger for heavier ion. easier to see if we note KE = P^2/(2m) , hence r is larger by approx sqrt (26/24) i would guess

Answer 3

0.40 5 m from the backside of the cylinder, if the spout is angled with the intention to allow the fluid to flow up vertically, this is. The rigidity of the top of water pushes the water as much as a point this is the comparable because of the fact the exterior of the cylinder. The diameters of the cylinder and that of the spout have no impression if the fluid is considered suitable (non viscous).