Electric Potential Difference for Speed of Light Percentage?

[aszty8]

Homework Statement

An X-ray tube is similar to a cathode-ray tube. Electrons are accelerated to high speeds at one end of the tube. If they are moving fast enough when they hit the screen at the other end, they give up their energy as X-rays (a form of nonvisible light).

(a.) Through what potential difference should electrons be accelerated so that their speed is 1.00% of the speed of light when they hit the other end of the tube?
(b.) What speed would this potential difference give to protons? Express your answer in m/s.
(c.) What potential difference would be needed to give protons the same kinetic energy as the electrons?

This is literally all we are told for this problem so I'm super confused. How am I supposed to go about solving this? Any help would be greatly appreciated. Thank you in advance.

Homework Equations

ΔV = ΔU/q₀ = -W_AB / q₀

The Attempt at a Solution

so if our change in speed is 1%, I think my $\Delta$V would be .01, but I have no idea what to do from there (or if I'm even on the right track.

 

[mfb]

so if our change in speed is 1%

Not a "change in speed". The initial speed is zero, and the final velocity is [what]?

I think my ΔV would be .01

No, that does not work. Just check the units: ΔV is an energy, 0.01 is a number.

 

[aszty8]

Oh god I feel so dumb now... The final velocity would be 1% of the speed of light, so 1% of 299792458 m/s, which would be 2997924.58 (or 2.99 x 10^6).

So our ΔV would be 2997924.58 ?

Am I being really dumb and mixing up velocity and volt?

 

[mfb]

So our ΔV would be 2997924.58 ?

Stop guessing, please.
What is ΔV? It is neither velocity nor Volt. How is that quantity related to the speed of electrons?

 

[aszty8]

I know that

V=U/q₀

And that

V_B-V_A=U_B/q₀-U_A/q₀=-W_AB/q₀
ΔV=ΔU/q₀=-W_AB/q₀

 

[mfb]

aszty8, my question is really basic (and I think your problem is exactly at that point): What type of physical quantity is V (and ΔV)? Is it a length? Is it a time? No, it is not, but do you know what it is?