-1
$\begingroup$

So I was recently thinking about what an electron's velocity would be after it was accelerated through a potential difference of 1 V. Doing simple calculations like just dividing its energy (which would be 1 eV after being accelerated) by m/2 (from the K.E equation) results in a velocity of about 592000 m/s which is totally absurd. I know that mass increases with velocity so the calculation that I did was wrong. So can anyone please tell me how to work out the actual velocity of the electron using, presumably, the concepts from SR?

Thank you

$\endgroup$
  • $\begingroup$ This is a simple application of $E^2=p^2c^2+m^2c^4$, combined with $p=\gamma m v$. $\endgroup$ – Danu Feb 20 '16 at 9:28
  • 2
    $\begingroup$ Yep, that's correct. What's absurd about it? An electron is very light and 592km/s is very slow compared to the speed of light. No relativity needed. $\endgroup$ – CuriousOne Feb 20 '16 at 9:54
  • $\begingroup$ I agree with CuriousOne. The speed of light would be exactly 299792458 m/s. Any speed considered as relativistic should be comparable to at least a twentieth of that speed. (exaggeration) $\endgroup$ – Horus Feb 20 '16 at 10:05
  • $\begingroup$ D'OH! -_- @Danu, CuriousOne and Horus, I overlooked the fact that light's speed was 300,000 km/s and not m/s! Anyway, the question remains. What would have happened if the potential difference had been 10^6 V instead of 1 V? Now, the velocity using the simple calculations comes out to be 592,999,453 m/s (cross checked it :p ). How would I tackle this now? $\endgroup$ – Anindya Mahajan Feb 20 '16 at 12:36
  • 2
    $\begingroup$ Please note that this is not a homework site. Most questions of homework-type (and this includes many questions that are not explicitly the coursework for some academic institution) or "check-my-work"-type are off-topic here. In particular, for your question to be well-received, it is important that you (1) show some effort of your own (2) ask a conceptual question. $\endgroup$ – Danu Feb 20 '16 at 13:05
1
$\begingroup$

As the comments have said, we deprecate questions that just involve churning through the maths without any conceptual content. However in this case I think there's room to discuss the concepts without things getting too homework like.

You started with the equation for the kinetic energy:

$$ E = \tfrac{1}{2} mv^2 \tag{1} $$

and knowing the energy (1eV) you worked out the corresponding velocity. This is fine as long as you bear in mind that equation (1) is an approximation that applies when $v \ll c$. As it happens this is fine for the 1ev energy you are considering, but you ask what happens when the energy is much higher. In that case you have to use an extended version of equation (1) that applies for high velocities:

$$ E^2 = p^2c^2 + m^2c^4 \tag{2} $$

where $p$ is the relativistic momentum:

$$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$

While it's easy to get the velocity from the energy using equation (1), doing this with equation (2) is slightly messier, but only slightly. Substituting for $p$ in equation (2) and rearranging we get:

$$ \frac{E^2}{c^2} - m^2c^2 = \frac{m^2v^2}{1 - \frac{v^2}{c^2}} $$

which simplifies to:

$$ v = c\sqrt{1 - \frac{m^2c^4}{E^2}} $$

The last wrinkle is that the energy $E$ in this equation is the total energy i.e. the 1eV of energy the electron got from the potential difference plus the rest mass energy given by Einstein's famous equation $E = mc^2$. So if $V$ is the voltage, so the potential energy is $eV$, the full equation is:

$$ v = c\sqrt{1 - \frac{m^2c^4}{(eV + mc^2)^2}} \tag{3} $$

Just for fun I graphed this up to a million volts to see what it looked like:

speed vs voltage

Incidentally, equation (3) gives the velocity for 1 volt as 593096 m/s compared to your result of (as you say) about 592000 m/s.

$\endgroup$
  • $\begingroup$ Oh, well thanks! I did actually calculate it afterwards and it came out around 299,999,998 m/s (used c=3*10^8 m/s to keep the calculations a bit simpler) $\endgroup$ – Anindya Mahajan Feb 20 '16 at 18:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.