Physics
7th Edition
ISBN: 9780321929013
Author: Douglas C. Giancoli
Publisher: PEARSON
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Chapter 17, Problem 12P
What is the speed of an electron with kinetic energy (a) 850 eV. and (b) 0.50 keV?
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What is the speed of an electron after being accelerated from rest through a 2.5×107 V potential difference?
Express your answer as a fraction of cc.
Suppose an electron (q= -e = -1.6 x 10^-19 C, m = 9.1 x 10^-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. Use the template in the attached photo to solve for the problem.
Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for
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the final speed of the electron. Express numerical answer in two significant figures.
The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation:
U =
Assuming all potential energy U is converted to kinetic energy K,
K+U = 0
K = -U
1
Since K
mv and using the formula for potential energy above, we arrive at an equation for speed:
2
v = (
1/2
Plugging in values, the value of the electron's speed is:
x 107 m/s
V=
Chapter 17 Solutions
Physics
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