The question I have is with respect to this diagram. Which depicts a stream of protons and electrons entering a proton field with equal energy. enter image description here

Why is it that in the case where the energy of protons and electrons are equal, their angle of deflection is exactly the same? However when speed is kept the same, the deflection of electrons is greater. Why is this so?

  • $\begingroup$ a better quality image would be appreciated and appreciated! In an electrostatic field, the force undergone is worth $\vec{F} = q \vec{E} $ therefore the force depends only on the charge of the particle $\endgroup$
    – Nicolas
    Mar 1, 2021 at 8:14
  • 9
    $\begingroup$ I’m voting to close this question because it's not about chemistry. $\endgroup$
    – Mithoron
    Mar 1, 2021 at 14:08
  • $\begingroup$ @Mithoron it's about mass spectrometry, isn't it? $\endgroup$ Mar 1, 2021 at 22:21
  • $\begingroup$ @LevelRiverSt: It's a question about basic electromagnetism, which'd commonly be accepted as a straightforward Physics topic. $\endgroup$
    – Nat
    Mar 2, 2021 at 12:47
  • $\begingroup$ @Nat it had not escaped my notice that was about basic electromagnetism. I was pointing out that OP had presumably posted it due to its application in mass spectrometry, an important technique in analytical chemistry. If so, I think it has a place here. $\endgroup$ Mar 2, 2021 at 14:23

1 Answer 1


Now that's a mildly non-trivial observation. Why would they be equal, really?

Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.

That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:

  1. Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
  2. Same energy: same $mv^2$, same deflection (just to the opposite sides).

Things get hairy in relativity, but that's another story.

So it goes.


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