Suppose the electric field between the electric plates in the mass spectrometer

Question

Suppose the electric field between the electric plates in the mass spectrometer of Figure 27-33 in the textbook is 2.41×104 V/m and the magnetic fields are B=B=0.36T. 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 1.66×1027kg.)

A. How far apart are the lines formed by the singly charged ions of each type on the photographic film? Express your answer using two significant figures. Enter your answers numerically separated by a comma.

B. What if the ions were doubly charged? Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Your help would be greatly appreciated!

Explanation / Answer

A) Velocity selection gives all ions the same velocity v
v = E/B = (2.41*10^4V/m) / (0.36T) =6.69*10^5 m/s

• Magnetic force provides the centripetal force ..
Bqv = mv²/R
R = mv / Bq ..

for C12 ... (m=12k,where k = 1.66*10^-27 kg)
> R(12) = (12k)v / Bq

for C13 .... m=13k
>R(13) = (13k)v / Bq

R(13) - R(12) =(kv / Bq)(13 - 12)
>R = kv / Bq
q = 1.6*10^-19 C
>R = {(1.66*10^-27)(6.69*10^5)} / {(0.36)(1.6*10^-19)} = 1.92*10^-2 m

R = difference in radii, on film it's a difference in diameters ..
sep. = 3.84*10^-2 m (3.84 cm)

B)   Doubling charge to 2q, doubles the magnetic force (vel.v unchanged due to velocity selection) so radius is

halved .. diameters become halved (sep. = 1.92 cm)

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