A scanning electron microscope uses a uniform 15.0-{/rm kN/C} electric field to

Question

A scanning electron microscope uses a uniform 15.0-{/rm kN/C} electric field to accelerate electrons horizontally toward the subject to be imaged. After travelling 5.0 {/rm cm} the electron are accelerated to a speed of 1.62*10^7 {/rm m/s}. The next step is to deflect the electrons so that they can scan across the sample- hence the scanning electron microscope. To accomplish this, the electrons are directed between a pair of oppositely charged parallel plates, which produce a uniform electric field of 6.42*10^3 {/rm N/C} perpendicular to the electron beam. This field extends for 3.5 {/rm cm} along the electron's path. For the scanning electron microscope described above, find the electron's vertical displacement.

Explanation / Answer

Solution: After the horizontal acceleration, the horizontal speed of electron is vx = 1.62*107 m/s

When electron enters the vertical field with this speed, it undergoes the vertical acceleration ay. It accelerates vertically while moving in horizontal direction over a horizontal distance of x = 3.5 cm = 0.035 m with the horizontal speed of vx = 1.62*107 m/s.

Hence the time taken by the electron to cover horizontal distance x = 0.035 m is given by

t = x/ vx

t = (0.035 m)/(1.62*107 m/s)

t = 2.1605*10-9 s

Initial vertical velocity of the electron is viy = 0 m/s

The vertical force on the electron due to the vertical electric field Ey = 6.42*103 N/C (from the oppositely charged parallel plates) is given by,

Fy = q*Ey

Fy = (1.6*10-19 C)*( 6.42*103 N/C)

Fy = 1.0272*10-15 N

Thus vertical acceleration due to this force is given by,

ay = Fy/me

ay = (1.0272*10-15 N)/(9.11*10-31 kg)

ay = 1.1276*1015 m/s2

Thus vertical distance travelled by the eletron is given by,

y = viy*t + (1/2)* ay*t2

y = 0 + (1/2)*( 1.1276*1015 m/s2)*( 2.1605*10-9 s)2

y = 2.6316*10-3 m

y = 0.26 cm

Thus electron's net vertical displacement is 0.26 cm

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