# Modern Physics Introduction Question on Electron Speed and Momentum

## 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:
$$1 eV = 1.602x10^{-19}J$$
mass of electron $$= 9.109x10^{-31}kg$$
c = $$2.998x10^{8}\frac{m}{s}$$

Functions:
Kinetic Energy(m,v)

$$KE=\frac{1}{2}mv^{2}$$

Momentum(m,v)

$$\vec{p}=m\vv}$$

## The Attempt at a Solution

Speed:

Kinetic Energy $$= 1.0eV = 1.602x10^{-19}J$$

$$KE=\frac{1}{2}mv^{2}$$

$$1.602x10^{-19}J = \frac{1}{2} (9.109x10^{-31}kg)(v)^2$$

$$3.204x10^{-19}J = (9.109x10^{-31}kg)(v)^{2}$$

$$\frac{3.204x10^{-19}J}{9.109x10^{-31}kg} = v^{2}$$

$$v^{2}=\frac{3.204x10^{-19}J}{9.109x10^{-31}kg}$$

$$v^{2}=3.517x10^{11}\frac{J}{kg}$$

$$v^{2}=3.517x10^{11}\frac{J}{kg}$$

$$\sqrt{v^{2}=3.517x10^{11}\frac{J}{kg}}$$

$$v=5.931x10^{5}\frac{m}{s}$$

So the electron is travelling $$5.931x10^{5}\frac{m}{s}$$ or 0.1978% of the speed of light

Momentum:

$$p=mv$$

$$p = (9.109x10^{-31}kg)(5.931x10^{5}\frac{m}{s})$$

$$p = 5.403x10^{-25} \frac{kg*m}{s}$$

Momentum is $$p = 5.403x10^{-25} \frac{kg*m}{s}$$