ALSO PART C!! C: What if the ions were doubly charged? ANSWER IN (R12 AND R13),(

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ALSO PART C!!

C: What if the ions were doubly charged?

ANSWER IN (R12 AND R13),(R13 AND R14)

Problem 20.57 Part A Suppose the electric field between the electric plates in the mass spectrometer of (Figure 1) is 2.98 × 104 V/m and the magnetic fields B-B' 0.65 T. 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 × 10-27 kg.) 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 and include the appropriate units IA r12 and 13 = 11.4. 10 Submit My Answers Give Up Figure 1 of 1 Incorrect; Try Again; 3 attempts remaining S2 Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures Sample B Part 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 and include the appropriate units Detector or film . . ris and Value Units

Explanation / Answer

Velocity selection gives all ions the same velocity v
v = E/B = (2.98*10^4V/m) / (0.65T) = 4.585*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 = 1.602*10^-19 C
>R = {(1.67*10^-27)(4.585*10^4)} / {(0.65)(1.602*10^-19)} = 7.353*10^-4 m

R = difference in radii, on film it's a difference in diameters ..
sep. = 14.705*10^-4 m (1.5*10^-3 m)

(2) sep. = 14.705*10^-4 m = 1.5*10^-3 m

(3) Doubling charge to 2q, doubles the magnetic force (vel.v unchanged due to velocity selection) so radius is halved .. diameters become halved (sep. = 7.4*10^-4 m)

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