Two isotopes of carbon, carbon-12 and carbon-13, have masses of 1.993 10-26 kg and 2.159 10-26 kg, respectively. These two isotopes are singly ionized (+e) and each is given a speed of 6.63 105 m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.7000 T. Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

Spatial separation = 0.0196m

Explanation:

From the question, we are given;

Mc12 = 1.993 x 10^(-26)

Mc13 = 2.159 x 10^(-26)

Velocity (v) = 6.663 x 10^(5) m/s

Charge on electron; q= 1.6 x 10^(-19)

Magnetic Field (B) = 0.7T

Formula for radius of circular paths for isotopes is;

r = Mv/qB

So applying it to this question,

r1 = M(c12)v/qB = [1.993 x 10^(-26) x 6.663 x 10^(5)] / [(1.6 x 10^(-19)) x 0.7]

= 0.1186

r2 = M(c13)v/qB = [2.159 x 10^(-26) x 6.663 x 10^(5)] / [(1.6 x 10^(-19)) x 0.7] = 0.1284

From the figure I attached, the spatial separation between the charged particles after half of the bigger circle is seen to be 2r2 - 2r1

Thus,

Spatial separation = 2r2 - 2r1 = (2 x 0.1284) - (2x0.1186) = 0.0196m