Question

In: Physics

Suppose the electric field between the electric plates in the mass spectrometer of (Figure 1) is...

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 ×10−27kg.)

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

Solutions

Expert Solution

________________________________________________________________

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

__________________________________________________________

(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

_____________________________________________________________________

(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

______________________________________________________________________

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

______________________________________________________________


Related Solutions

Suppose the electric field between the electric plates in the mass spectrometer of Figure 27-33 in...
Suppose the electric field between the electric plates in the mass spectrometer of Figure 27-33 in the textbook is 2.64×104 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 1.66×10−27kg.) How far apart are the lines formed by the singly charged ions of each type on the photographic film? What if the ions were doubly charged?
Suppose the electric field between the electric plates in the mass spectrometer of Fig. 20-38 is...
Suppose the electric field between the electric plates in the mass spectrometer of Fig. 20-38 is 2.38 ? 104 V/m and the magnetic fields B = B' = 0.66 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.) A) How far apart are the lines formed by the singly charged ions of each type on the photographic film? mm...
Consider the mass spectrometer shown schematically in the figure below. The magnitude of the electric field...
Consider the mass spectrometer shown schematically in the figure below. The magnitude of the electric field between the plates of the velocity selector is 2.40 103 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.040 0 T. Calculate the radius of the path for a singly charged ion having a mass m = 2.04 10-26 kg.
1) A parallel plate capacitor is connected to a battery. The electric field between the plates...
1) A parallel plate capacitor is connected to a battery. The electric field between the plates is E. While still connected to the battery, we move the plates so that their plate separation is now twice as large. What is the electric field between the plates now? E/4. E. E/2. 4E. 2E. 2) A parallel plate capacitor with capacitance Co is fully charged. The plates are in the shape of a disk. If the diameter of the disk is doubled...
What is the magnitude of the electric field at the dot in the figure? (Figure 1)
Part A What is the magnitude of the electric field at the dot in the figure? (Figure 1) Express your answer using two significant figures. E = _________V/m   Part B  What is the direction of the electric field at the dot in the figure? Choose the best answer. (a) the negative x-axis. (b) the positive x-axis. (c) 45 below -x-axis (d) 45 below +x-axis
What is the direction of the electric field at the dot in the figure (Figure 1) ?
Part AWhat is the direction of the electric field at the dot in the figure (Figure 1) ?a) to the rightb) downc) upd) to the leftPart BWhat is the magnitude of the electric field at the dot?Express your answer to one significant figure and include the appropriate units.
Determine the magnitude and direction of the electric field at point 1 in the figure(Figure 1).
Figure 1 Part A Determine the magnitude and direction of the electric field at point 1 in the figure(Figure 1). E1→=(2500V/m,up) E1→=(7500V/m,up) E1→=(3750V/m,down) E1→=(2500V/m,down) Part B Determine the magnitude and direction of the electric field at point 2 in the figure. E2→=(2500V/m,up) E2→=(3750V/m,down) E2→=(7500V/m,down) E2→=(5000V/m,up)
The electric field between the plates of a paper-separated (K=3.75) capacitor is 8.28
The electric field between the plates of a paper-separated (K=3.75) capacitor is 8.28
you have a parallel-plate capacitor a.) determine the electric field between the plates if there is...
you have a parallel-plate capacitor a.) determine the electric field between the plates if there is a 120 V potential dofference across the plated and they are separated by 0.5 cm b.) A spark will jump if the magnitude of the electrix field between the plates exceeds 3.0x10^6 V/m when air separates the plates. what is the closest the plated can be place to esch ither without sparking c.) if a dialectric (k=2.5) is inserted between the plates how will...
The electric field between two parallel plates is uniform, with magnitude 576 N/C. A proton is...
The electric field between two parallel plates is uniform, with magnitude 576 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.06 cm. At the same moment, both particles are released. (a) Calculate the distance (in cm) from the positive plate at which the two pass each other. Ignore the electrical attraction between the proton and electron. _______cm (b) Repeat part (a) for a sodium...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT