Speed of Free Electron Moving in Space

[quickclick330]

The total energy of a free electron moving through empty space is E=1.5mc^2, where m is the mass of the electron and c is the speed of light. What is this electron’s speed?

1. c (the speed of light)
2. 0.7453c
3. 0.8660c
4. 0.9428c
5. 0.9950c
6. 0.9999c


I thought since that Etotal = 1.5mc^2 then...

Etotal = KE = .5mv^2

1.5mc^2 = .5mv^2 and solve for v (masses would cancel which is what I was looking for)

What is wrong about this approach?

 

[Doc Al]

Using KE = .5mv^2 is only good when v << c.

Hint:
$$E = \gamma m c^2$$

 

[kudoushinichi88]

I believe that the proper relativistic kinetic energy equation would be

$$E_k = (\gamma - 1)mc^2$$

Or did you make that mistake on purpose, Doc Al?

 

[quickclick330]

oh shoot...i forgot about that. Do I need to use the non-approximated version of KE? (I can never remember it.. but I'm looking it up right now)

 

[Doc Al]

E is total energy, of course.

(There's no need to find kinetic energy.)

 

[quickclick330]

okay

 

[quickclick330]

so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?

 

[Doc Al]

so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?

Yep. That's all you need to do.

 

[quickclick330]

ahh...got it, thank you :-)

 

[kudoushinichi88]

Oh, I get it... the total energy of the free electron is the sum of its kinetic energy and the energy from the mass-energy equivalence? The potential energy of the electron is zero, since it's free, right?

$$E_{total}= (\gamma - 1)mc^2 + mc^2$$
$$\ \ \ = \gamma mc^2$$

Is this so?

 

[Doc Al]

Oh, I get it... the total energy of the free electron is the sum of its kinetic energy and the energy from the mass-energy equivalence? The potential energy of the electron is zero, since it's free, right?

$$E_{total}= (\gamma - 1)mc^2 + mc^2$$
$$\ \ \ = \gamma mc^2$$

Is this so?

Yep.

 

[kudoushinichi88]

Ah, thanks...