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When learning about X-rays we are told that they are created through braking radiation. When we are learning about braking radiation we are told that this occurs when an electron passes close to a nucleus and slows down causing a reduction in kinetic energy, which is then emitted as an X-ray.

The problem that I have is why it slows down. The electron would be accelerated towards the nucleus as it approaches and then I would expect it to be negatively accelerated by the same amount as it travels away from the nucleus. Thus, I would expect to see the kinetic energy conserved over the whole event and no braking radiation. What am I missing here?

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  • $\begingroup$ Where did you read that explanation? $\endgroup$
    – KF Gauss
    Commented Sep 27, 2022 at 12:33
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    $\begingroup$ It is the whole atom with which an electron interacts, and the nucleus is shielded by the orbitals of the electrons all about the nucleus. The field between electrons and electron orbitals is repulsive, not attractive. $\endgroup$
    – anna v
    Commented Sep 27, 2022 at 12:46
  • $\begingroup$ @annav, thank you. So you wouldn't expect them to be equally shielded on the way out? $\endgroup$
    – James
    Commented Sep 27, 2022 at 13:06
  • $\begingroup$ they will be continually interacting with the field of the orbital electr[ons, in and out. $\endgroup$
    – anna v
    Commented Sep 27, 2022 at 13:27
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    $\begingroup$ @James The interaction of electron with the atom's electron clouds can have both elastic and non-elastic component. The argument for equal shielding in the 2nd paragraph works only for elastic scattering, which is not the case here. $\endgroup$
    – KP99
    Commented Sep 27, 2022 at 16:38

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The explanation you've given is a bit mangled, and it's not surprising that (under the assumptions you state) you've come to the conclusion that there should be no effect.

The key piece that you're missing is that according to classical electrodynamics, accelerating charges emit electromagnetic waves. These waves carry energy away to infinity, and so (by conservation of energy) the total energy of the accelerating charge must decrease. The intensity of the emitted waves is proportional to the square of the acceleration of the charge, and so there's no cancellation between the "ingoing" and "outgoing" travel of the electron; both legs of the trip emit a positive amount of energy. These emitted waves are what we observe as bremsstrahlung. (The details get more complicated when we add in quantum mechanics, but that's not really necessary to get a picture of why the electron's energy decreases.)

In the limit of "small" acceleration, the radiated energy goes to zero, and so we can often neglect the effects of this energy loss. In this limit, your logic would be correct: the particle would have a certain amount of KE when it was far away from the scattering atom, would accelerate and change its KE as it approached the atom, and then would return to its original amount of KE as it flew away from the atom. It's only because the acceleration of the electron is sufficiently large that this "energy loss to radiation" effect becomes important.

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    $\begingroup$ One of these days I will spell "bremsstrahlung" correctly on the first try. Today is not that day. $\endgroup$
    – Michael Seifert
    Commented Sep 27, 2022 at 13:17
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    $\begingroup$ An aid? brems - - - strah - - - lung :-) $\endgroup$
    – Farcher
    Commented Sep 27, 2022 at 13:25
  • $\begingroup$ @KP99: I'm not sure I understand your point. The OP is asking why there is a net change in the KE of the electron between when it is "ingoing" and "outgoing" (but very far away in both cases.) See the second-to-last sentence. Certainly it's true that the electron's kinetic energy is different when it's closer to the atom/nucleus compared to when it's far away, but I don't think the OP is asking about that. $\endgroup$
    – Michael Seifert
    Commented Sep 27, 2022 at 17:21
  • $\begingroup$ @Farcher: My issue is that I always think it's "brehms" rather than "brems". $\endgroup$
    – Michael Seifert
    Commented Sep 27, 2022 at 17:21
  • $\begingroup$ Ah, my bad! I wasn't sure if OP was asking the far away case for both 'incoming' and 'outgoing' electron. It makes sense now, thanks for the clarification! +1 $\endgroup$
    – KP99
    Commented Sep 28, 2022 at 8:58

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