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 1) is 3.18×104V/m and the magnetic fields B=B=0.65T. 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 ×1027kg.) A: How far apart are the lines formed by the singly charged ions of mass numbers 12 and 13 on the photographic film? B:How far apart are the lines formed by the singly charged ions of mass numbers 13 and 14 on the photographic film?

Explanation / Answer

According to the given problem,

Velocity selection gives all ions the same velocity v
v = E/B = (3.18*10^4V/m) / (0.65T) =4.89*10^4 m/s

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

for C12 ... (m=12k,where k = 1.67*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 = {(1.67*10^-27)(4.89*10^4)} / {(0.65)(1.6*10^-19)} = 7.856*10^-4 m

R = difference in radii, on film it's a difference in diameters ..
sep. = 1.57*10^-3 m (1.57mm)

(2) repeat for C13 and C14 .... (same answer)

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