Question

Suppose the electric field between the electric plates in the mass spectrometer of the figure 20-39 in the textbook is 2.69×104V/m and the magnetic fields B=B′=0.70T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67 ×10−27kg.)

A - How far apart are the lines formed by the singly charged ions of mass numbers 12 and 13 on the photographic film?

Express your answer using two significant figures.

r12 and 13 = ? m

B- How far apart are the lines formed by the singly charged ions of mass numbers 13 and 14 on the photographic film?

Express your answer using two significant figures.

r13 and 14 = ? m

C- What if the ions were doubly charged?

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

r12 and 13, r13 and 14 = ? m

Answer 1

The electric field,E= 2.69×10⁴ V/m

The magnetic fields B=0.70 T

All the ions have same velocity
velocity ,v = E/B = (2.69*10⁴ V/m) / (0.07T) =3.84*10⁵ m/s

We know that the Magnetic force provides the centripetal force, So equate the magnetic force and the centripetal force, thus
Bqv = mv²/R
R = mv / Bq
For C¹² (mass,m=12k,where k = 1.67*10^-27 kg)
=> R(12) = (12k)v / Bq

for C^13, mass m=13k
=>R(13) = (13k)v / Bq

So the difference, R(13) - R(12) =(kv / Bq)(13 - 12)
=> or R = kv / Bq
q = 1.6*10^-19 C
So, R = {(1.67*10^-27)(3.84*10⁵)} / {(0.07)(1.6*10^-19)} = 0.0573 m

where R = difference in radii, on film it's a difference in diameters.
Diameter = 0.114 m

(2)
For C¹³ (mass,m=13k,where k = 1.67*10^-27 kg)
=> R(13) = (13k)v / Bq

for C^14, mass m=14k
=>R(14) = (14k)v / Bq

So the difference, R(14) - R(13) =(kv / Bq)(14 - 13)
=> or R = kv / Bq
q = 1.6*10^-19 C
So, R = {(1.67*10^-27)(3.84*10⁵)} / {(0.07)(1.6*10^-19)} = 0.0573 m

where R = difference in radii, on film it's a difference in diameters.
Diameter = 0.114 m
C- Doubling charge to 2q, which will doubles the magnetic force , that makes the radius halved and the diameters become halved ,so radius,r=0.0573 m