Answer:
R' = 4R
Explanation:
You have that the radius of curvature for a ion of mass m, charge q and speed v, is R.The ion is moving in a zone with a magnetic field. You take into account the following formula for the radius of the trajectory of the ion:
[tex]R=\frac{mv}{qB}[/tex] (1)
m: mass of the ion
q: charge of the ion
B: magnitude of the magnetic field
v: speed of the ion
If another ion of mass 2m, charge q and speed 2v enters to the same magnetic field, you use the equation (1) to obtain the new radius of curvature:
[tex]R'=\frac{(2m)(2v)}{qB}=4\frac{mv}{qB}=4R[/tex]
Then, the new radius of curvature is four times the radius of curvature of the ion with mass m
