Question: In a TV picture tube the accelerating voltage is 15.0kV and the electron beam passes through an aperture 0.50 mm in diameter to a screen

In a TV picture tube the accelerating voltage is 15.0kV and the electron beam passes through an aperture 0.50 mm in diameter to a screen 0.300 m away.
(a) Calculate the uncertainty in the component of the electron's velocity perpendicular to the line between aperture and screen.
(b) What is the uncertainty in position of the point where the electrons strike the screen?
(c) Does this uncertainty affect the clarity of the picture significantly? (Use non-relativistic expressions for the motion of the electrons. This is fairly accurate and is certainly adequate for obtaining an estimate of uncertainty effects.)

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