# 2.2 eV electron

## Homework Statement

I need to figure out the energy, rest mass, speed, wavelength and momentum of a 2.2 eV electron. Thats all it says.

Ek = 1/2mv^2

## The Attempt at a Solution

I dont understand what 2.2 eV is. The kinetic energy?
This converts to 3.52x10^-19 J

Rest mass should be 9.11x10^-31. Or do I use E=mc^2? Why or why not?

No idea how to find speed, which I need to calculate wavelength and momentum. I could use Ek= 1/2mv^2 because the electron will not be travelling close to the speed of light, but I feel like this is not the way you SHOULD calculate it. There has to be another way.

gneill
Mentor

## Homework Statement

I need to figure out the energy, rest mass, speed, wavelength and momentum of a 2.2 eV electron. Thats all it says.

Ek = 1/2mv^2

## The Attempt at a Solution

I dont understand what 2.2 eV is. The kinetic energy?
This converts to 3.52x10^-19 J
Yes, that'll be the kinetic energy.
Rest mass should be 9.11x10^-31. Or do I use E=mc^2? Why or why not?

No idea how to find speed, which I need to calculate wavelength and momentum. I could use Ek= 1/2mv^2 because the electron will not be travelling close to the speed of light, but I feel like this is not the way you SHOULD calculate it. There has to be another way.

Check the speed that you get when you apply Newtonian physics to find the speed from the KE . If its much less than the speed of light then you can continue with Newtonian physics and ignore relativistic effects. So, what value do you get for speed via 1/2mv2?

8.79x10^5 m/s

gneill
Mentor
8.79x10^5 m/s
How does that compare to the speed of light (percentage-wise)?

Its 0.293% of the speed of light. But I'm mainly concerned about how to do the question without Ek=1/2mv^2. What if it WAS close to the speed of light? Is there a different procedure I can use?

gneill
Mentor
Its 0.293% of the speed of light. But I'm mainly concerned about how to do the question without Ek=1/2mv^2. What if it WAS close to the speed of light? Is there a different procedure I can use?
Yes, you would need to invoke the formulas of special relativity to deal with how some of the energy ends up as an apparent increase in mass. Note though that the rest mass is always the same regardless. Rest mass is the mass of an object that is at rest with respect to the observer, so no corrections for speed are involved at all.

Can you outline the procedure? I really don't understand this last unit of quantum mechanics. Like what is the full equation? I would be solving for v

gneill
Mentor
Can you outline the procedure? I really don't understand this last unit of quantum mechanics. Like what is the full equation? I would be solving for v
Well, the relativistic kinetic energy is given by:

$$KE = m_o c^2 \left[ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right]$$

and you would be looking for v, which will get a tad messy algebraically.