Part AAn electron is to be accelerated from a velocity of 1.50x106 m/s to a velocity of 9.00×106 m/s. Through what potential difference must the electron pass to accomplish this?Express your answer in volts.V₁ V₂ = -224 V-SubmitPrevious AnswersCorrectPart BThrough what potential difference must the electron pass if it is to be slowed from 9.00x106 m/s to a halt?Express your answer in volts.ΠΑΣΦV₁ V₂ = 172.76?V

Part A. Mass of electron = m=9.1×10-31 Kg Charge on electron = q=1.6×10-19 C Initial velocity = u =…

Answered: An electron is to be accelerated from a…

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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Part A
An electron is to be accelerated from a velocity of 1.50x106 m/s to a velocity of 9.00×106 m/s. Through what potential difference must the electron pass to accomplish this?
Express your answer in volts.
V₁ V₂ = -224 V
-
Submit
Previous Answers
Correct
Part B
Through what potential difference must the electron pass if it is to be slowed from 9.00x106 m/s to a halt?
Express your answer in volts.
ΠΑΣΦ
V₁ V₂ = 172.76
?
V

Expert Solution

Step 1

Part A.

Mass of electron = $m = 9.1 \times 10^{- 31} Kg$

Charge on electron = $q = 1.6 \times 10^{- 19} C$

Initial velocity = u = $1.5 \times 10^{6} m / s$

Final velocity = v = $9.0 \times 10^{6} m / s$

According to work energy theorem:

Work done = Change in kinetic energy.

$q (V_{1} - V_{2}) = \frac{1}{2} m (v^{2} - u^{2}) (V_{1} - V_{2}) = \frac{1}{2} \frac{m (v^{2} - u^{2})}{q} (V_{1} - V_{2}) = \frac{1 \times 9.1 \times 10^{- 31} [{(9 \times 10^{6})}^{2} - {(1.5 \times 10^{6})}^{2}]}{2 \times (- 1.6 \times 10^{- 19})} (V_{1} - V_{2}) = - 224 \times 10^{- 19} V$

Hence, potential difference is $- 224 V$ .

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