Question

Electrons are accelerated through a voltage difference of 280 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
which I found to be 2.8E5eV

Answer 1

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

Answer 2

The answer to your first part is correct, I believe. Should just be K = qV where q = e.

For the speed of the electrons we will consider relativistic Kinetic Energy. Kinetic Energy in this realm of physics is considered to be the relativistic energy of the electron minus the rest energy of the electron. You might recall 'gamma' as being the constant for the relationships in relativistic circumstances.

kinetic energy K = 2.8 x 10^5 eV = 0.28 MeV
rest mass Mo = 0.511 MeV/c^2
relativistic mass M = γMo
speed of light c = 3 x 10^8 m/s
gamma γ = Contains the velocity in terms of 'c' we're looking for, see below

I'm just going to work out the algebra:\[K = mc^{2} - m_{o}c^{2}\]\[K = \gamma m_{o}c^{2}-m_{o}c^{2}\]\[K = (m_{o}c^{2})(\gamma - 1)\]\[\frac{ K }{ m_{o}c^2 }=\gamma - 1\]\[\frac{ K }{ m_{o}c^2 }+1 = \gamma = \frac{ 1 }{\sqrt{1-\beta^{2}}} ; \beta=\frac{ v^{2} }{ c^{2} }\]\[\sqrt{1 - \beta^{2}}(\frac{ K }{ m_{o}c^{2} + 1 }) = 1\]\[\sqrt{1 - \beta^{2}}= \frac{ 1 }{ \frac{ K }{ m_{o}c^{2} }+1 }\]\[1-\beta^{2}=(\frac{ 1 }{ \frac{ K }{ m_{o}c^{2} }+1 })^{2}=\frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }\]\[-\beta^{2} = \frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }-1\]\[\beta^{2} = 1 - \frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }\]\[\frac{ v^{2} }{ c^{2} } = 1 - \frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }\]\[\frac{ v }{ c } =\sqrt{1-\frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }}\]Hence 'v' in terms of 'c' is given by:\[v =c \sqrt{1-\frac{ 1 }{ (\frac{ K }{ m_{o}c^{2} }+1)^{2} }}\]
I'm not going to substitute values, it's simple. Just note that Mo is in terms of c^2 so that immediately cancels. You're really only doing 0.28 / 0.511 for that fraction.

I get a solution of: v = 0.763c