Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collidewith a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider onesuch electron that is emitted with an initial velocity of 1.90 x 107 m/s in the horizontal direction when magnetic forces deflect theelectron with a vertically upward acceleration of 5.50 x 1015 m/s2. The phosphorescent screen is a horizontal distance of 7.2 cmaway from the point where the electron is emitted.(a) How much time does the electron take to travel from the emission point to the screen?(b) How far does the electron travel vertically before it hits the screen?cm

Answered: Older televisions display a picture…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide
with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one
such electron that is emitted with an initial velocity of 1.90 x 107 m/s in the horizontal direction when magnetic forces deflect the
electron with a vertically upward acceleration of 5.50 x 1015 m/s2. The phosphorescent screen is a horizontal distance of 7.2 cm
away from the point where the electron is emitted.
(a) How much time does the electron take to travel from the emission point to the screen?
(b) How far does the electron travel vertically before it hits the screen?
cm

Expert Solution

Step 1

As acceleration is in vertical direction , it will not contribute to horizontal velocity : So:

$s = v t h e r e, s i s t h e d i s p l a c e m e n t v i s t h e v e l o c i t y a n d t i s t h e t i m e t h e n a s v = 1.9 \times 10^{7} m / s s = 7.2 c m = 0.072 m t h e n t = \frac{s}{v} = \frac{0.072}{1.9 \times 10^{7}} = 37.89 \times 10^{- 10} s$

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