In a mass spectrometer, a singly ionized 24Mg ion has a mass equal to 3.983 10-26 kg and is accelerated through a 2.00-kV potential difference. It then enters a region where it is deflected by a magnetic field of 537 G.
a. Find the radius of the curvature of the ion's orbit.
b. What is the difference in the orbital radii of the 26Mg and 24Mg ions? Assume that their mass ratio is 26:24

R = 0.584m

Explanation:

Given B = 537G = 537×10-⁴T = 0.054 T

Mass, m = 3.98×10-²⁶ kg, V = 2.00kV = 2000V. q = 1.6×10-¹⁹ C

The radius R of curvature is given by

R = mv/qB

We don't know the value for v the velocity, but we can get the expression for it from energy concepts.

The electric potential energy of the field is equal to the kinetic energy of the ion

So

qV = 1/2mv²

v² = 2qV/m = 2 ×1.6×10-¹⁹×2000 /(3.98×10-²⁶)

v² = 1.608×10¹⁰

v = sqrt(1.608×10¹⁰)= 126809m/s

So R = 3.98×10-²⁶ × 126809/(1.6×10-¹⁹ × 0.054)

R = 0.584 m

the area between the graph and the t axis

wavelength divides in half and speed stays the same.