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 2.08 times 10^4 V/m and the magnetic fields B = B' = 0.71 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 times 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. 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. What if the ions were doubly charged? Express your answer using two significant figures separated by a comma.

Explanation / Answer

to pass undeflected,

Fe - Fb = 0

q E - q v B = 0

v = E/B = (2.08 x 10^4) / (0.71) = 29295.77 m/s

this is same for both.(independent of mass)

after that for circular path,

Fb = m v^2 / r

q v B = m v^2 / r

r = m v / q B

r0 = m0 v / q B = (1.67 x 10^-27)(29295.77) / (1.6 x 10^-19 x 0.71)

= 4.307 x 10^-4 m

r12 = 12r0 = 5.168 x 10^-3 m

r13 = 5.599 x 10^-3 m

r14 = 6.029 x 10^-3 m

(A) 2(r13 - r12 ) = 8.6 x 10^-4 m

(B) 2(r14 - r13) = 8.6 x 10^-4 m

(c) if q -> 2q

then these distances will be halved.

r12 and 13 = 4.3 x 10^-4 m

r13 and 14 = 4.3 x 10^-4 m

          

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