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 2.58×104V/m and the magnetic fields B=B=0.77T. 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.)

Part 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 and include the appropriate units.

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.

Part C

What if the ions were doubly charged?

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

Screenshot for reference http://gyazo.com/50522858a819b1f61fa608c15f3e3ec8

Explanation / Answer

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Given,

E = 2.58 x 104 V/m ; B = 0.77 T

F(elect) = q E and F(mag) = q v B

we get, v = E / B

v = 2.58 x 104 / 0.77 = 3.35 x 104 m/s

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(A) the location of each line will be double the radius of curvature

We know that , F(mag) = q v B and F(cent) = Mv2/R this gives us

R = m v / q B ( q = e(charge on electron) = 1.6 x 10-19 C)

R(12) = 12 x 1.67 x 10-27 x  3.35 x 104 / 1.6 x 10-19 x 0.77 = 54.49 x 10-4 m

R(13) = 13 x1.67 x 10-27 x  3.35 x 104 / 1.6 x 10-19 x 0.77 = 59.03 x 10-4 m

R(14) = 14 x 1.67 x 10-27 x  3.35 x 104 / 1.6 x 10-19 x 0.77 = 63.57 x 10-4 m

R(13) - R(12) =2(  59.03 x 10-4 m - 54.49 x 10-4 m) =9.08 x 10-4 m

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(B) R(14) - R(13) =2(  63.57 x 10-4 m - 59.03 x 10-4 m) =9.08 x 10-4 m

They are seperated by R = 9.08 x 10-4 m

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Part(C)

If the ions are double charged then the radis and hence the seperation reduces by half.

R = m v / 2 e B

R(12) = 12 x 1.67 x 10-27 x  3.35 x 104 / 2 x1.6 x 10-19 x 0.77 = 27.25 x 10-4 m

R(13) = 13 x1.67 x 10-27 x  3.35 x 104 / 2 x1.6 x 10-19 x 0.77 = 29.52 x 10-4 m

R(14) =  14 x 1.67 x 10-27 x  3.35 x 104 / 2 x1.6 x 10-19 x 0.77 = 31.79 x 10-4 m

R(13) - R(12) =  29.52 x 10-4 m - 27.2 x 10-4 m =2.27  x 10-4 m

R(14) - R(13) =31.79 x 10-4 m - 29.52 x 10-4 m = 2.27  x 10-4 m

Hence the dist is = 2 x 2.27  x 10-4 m = 4.54 x 10-4 m

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