Question

I ONLY need help with B. Please show work. Thank you!

A) Electrons are accelerated through a voltage difference of 275 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?

B) What is the speed of these electrons in terms of the speed of the light? (Remember that the electrons will be relativistic.)?

Answer 1

I have a doubt, but here's my attemp and I hope it can help you. I solve part a) to be sure too, cause you are not providing the value obtained:

Potential difference V = 275000 V
Kinetic energy of the elecrons K = Vq
Where q = charge of electron
             = 1.6 x 10 ^-19 C

Substitute values we get K = 4.4 x 10 ^-14 J

(b).Speed of electron v = ?

We know K = m ' c ² - m c ²

Where m = mass of electron

               = 9.11 x 10 ^-31 kg

          m ' = relativistic mass

                = m / √[1-(v/c) ² ]

K / c ² = m ' - m

          = [m / √[1-(v/c) ² ]] - m

         = m {[ 1 / √[1-(v/c) ² ] - 1}

K / mc ² = {[ 1 / √[1-(v/c) ² ] - 1}

0.5366 = {[ 1 / √[1-(v/c) ² ] - 1}

{[ 1 / √[1-(v/c) ² ] } = 1.5366

√[1-(v/c) ² ] = 0.65

[1-(v/c) ² ] =0.4234

              v = 0.759 c

Hope it helps.