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fallen186
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Homework Statement
Suppose that an electron has a kinetic energy of 1.0 eV. What are the speed and the momentum of the electron?
The book is Modern Physics for Scientists and Engineers by Morrison. If anyone knows where to get any of the answers(or solutions. solutions may be harder to find) I would greatly appreciate it!
Homework Equations
Constants:
[tex]1 eV = 1.602x10^{-19}J[/tex]
mass of electron [tex]= 9.109x10^{-31}kg[/tex]
c = [tex]2.998x10^{8}\frac{m}{s}[/tex]
Functions:
Kinetic Energy(m,v)
[tex]KE=\frac{1}{2}mv^{2}[/tex]
Momentum(m,v)
[tex]\vec{p}=m\vv}[/tex]
The Attempt at a Solution
Speed:
Kinetic Energy [tex]= 1.0eV = 1.602x10^{-19}J[/tex]
[tex]KE=\frac{1}{2}mv^{2}[/tex]
[tex]1.602x10^{-19}J = \frac{1}{2} (9.109x10^{-31}kg)(v)^2[/tex]
[tex]3.204x10^{-19}J = (9.109x10^{-31}kg)(v)^{2}[/tex]
[tex]\frac{3.204x10^{-19}J}{9.109x10^{-31}kg} = v^{2}[/tex]
[tex]v^{2}=\frac{3.204x10^{-19}J}{9.109x10^{-31}kg} [/tex]
[tex]v^{2}=3.517x10^{11}\frac{J}{kg}[/tex]
[tex]v^{2}=3.517x10^{11}\frac{J}{kg}[/tex]
[tex]\sqrt{v^{2}=3.517x10^{11}\frac{J}{kg}}[/tex]
[tex]v=5.931x10^{5}\frac{m}{s}[/tex]
So the electron is traveling [tex]5.931x10^{5}\frac{m}{s}[/tex] or 0.1978% of the speed of light
Momentum:
[tex]p=mv[/tex]
[tex]p = (9.109x10^{-31}kg)(5.931x10^{5}\frac{m}{s})[/tex]
[tex]p = 5.403x10^{-25} \frac{kg*m}{s}[/tex]
Momentum is [tex]p = 5.403x10^{-25} \frac{kg*m}{s}[/tex]