As an example of the manipulation of an electron beam, consider an electron trav
ID: 251213 • Letter: A
Question
As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity vi = vii. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axi + ayj, where ax and ay are constants. For the case vi = 1.86 107 m/s, ax = 7.98 1014 m/s2, and ay = 1.55 1015 m/s2, determine the following, at x = d = 0.0100 m.
(a) the position of the electron yf = _____________
(b) the velocity of the electron vf = __________ m/s i + _________ m/s j
(c) the speed of the electron |vf| = ________ m/s
(d) the direction of travel of the electron (i.e. the angle between its velocity and the x axis) = ______°
Explanation / Answer
In the x direction v = vi + (ax)t
s = (vi)t + 0.5(ax)t2
0.01 = 1.95x107t + 0.5(8.04x10^14)t2
t = [-1.95x107 + sqrt(1.95x107)2 + 0.04*0.5(8.04x1014))]/[2(8.04x1014)]
= 2.54x 10(-10) s
Put that into y direction formula:
y = 0t + .5(1.52x1015)2
= 1.155
Don't bother about the angle. They give acceleration in the x and y directions already. Don't use g for acceleration.
(b)vf = {vi + (ax)t}ihat + {(ay)t}jhat
{1.95 +(8.04x 1014)}ihat+(8.04x 1.155x 1030)jhat
=(1.95+8.40x 1014)ihat +(9.28x 030)jhat
(c) use pythagoras on answers from part (b) Yf=1/2ayt^2
=1/2(.04x 1.155x 1030)2
= 5.33 x 1064
(d) for angle tan vf
i.e. tan((1.95+8.40x 1014)ihat +(9.28x 030)jhat)
Please refer to the following solve question, calculation may be differ but concept is rignt.
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