Question

Two isotopes of carbon, C12 and C13, have masses of 19.93x 10 kg and 21.59x1027 kg, respectively. These two isotopes are singly ionized (+e 1.602 x 1019 C) and each is given a speed of 6.828 x 10% m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.8878 T. Calculate the spatial separation, Ax (indicated by the double arrows), between the two isotopes after they have traveled through a half-circle. [Note: The magnetic force is equal to the centripetal force on the ions] B (out of screen) 丄. r12

Answer 1

Let the magnetic field B = 0.8878 T

Let the speed of the ions be v = 6.828 × 10⁵ m/s

Let the mass of isotope C12 be m1 = 19.93 × 10^-27 kg

Let the mass of the isotope C13 be m2 = 21.59 × 10^-27 kg

We know that charge e = 1.6 × 10^-19 C

Due to the effect of magnetic field, ions trace circular paths.

If we know the separation between these paths, we simply know the spatial separation between the two isotopes

Radius of a particle tracing a circular path under a magnetic field is given by r = m v / qB .............. ( 1 )

We can observe that larger the mass , larger the radius.

Let the diameter of the circle traced by the isotope C12 be D₁₂

Similarly the diameter of the circle traced by C13 be D₁₃

So the spatial separation between the two isotopes is

∆x = D₁₃ - D₁₂

= 2r₁₃ - 2 r₁₂

= 2 ( r₁₃ - r₁₂ )

Using equation ( 1 ) , we can write ,

∆x = 2 { ( m2 v / e B ) - ( m1 v / eB )}

= ( 2 v / eB ) × ( m2 - m1 )

= {( 2 × 6.828 × 10⁵ ) /( 1.6 × 10^-19 × 0.8878 ) } × ( 21.59 × 10^-27 - 19.93 × 10^-27 )

= ( 22.66 × 10^-22 ) / ( 1.42 × 10^-19 )

= 15.9 × 10^-3 m

= 0.0159 m

= 15.9 mm

So the separation between the two isotopes is ∆x = 15.9 mm