In X-ray diffraction when we accelerate the electron with some potential say $V$ and make collision with plate that produces $X-$ray (frequency $\nu$). $$h\nu_{max}=eV$$ But as we know that the accelearated charge particle radiate electromagnetic radiation which have energy. Why we don't take it into account in our calculations?


One can do a rough non-relativistic calculation to compare the energy radiated, $E_{\rm radiation}$, and the kinetic energy gained, $E_{\rm kinetic}$, by an electron when traversing a distance $d$ which has an accelerating potential difference $V$ across it.

The electric field is assumed to be uniform $E = \dfrac Vd$ and the transit time for an electron being $t$.

$E_{\rm kinetic} = \frac 12 m \,a^2\,t^2$ where $a$ is the acceleration of the electron.

The Larmor formula states that the power of the emitted radiation from an accelerating electron, charge $e$ and mass $m$, is $P = \dfrac{e^2\,a^2}{6\,\pi\,\epsilon_0\,c^3}$ where $c$ is the speed of light.

With $E_{\rm radiation} = \dfrac{e^2\,a^2\,t}{6\,\pi\,\epsilon_0\,c^3} \Rightarrow \dfrac{E_{\rm radiation}}{E_{\rm kinetic}} = \dfrac{e^2}{3\,\pi\, \epsilon_0\,c^3 \,m\,t}$.

$s= ut+\dfrac 12 at^2 \Rightarrow d = \dfrac 12 \dfrac {e\,V}{m\,d}\,t^2 \Rightarrow t = \sqrt{\dfrac{2\,m\,d^2}{e\,V}}$

With $V= 30 \,\rm kV$ and $d=0.1\,\rm m$ the transit time $t \approx 2\times 10^{-9}\,\rm s$.

Putting in the values relating to an electron gives $\dfrac{E_{\rm radiation}}{E_{\rm kinetic}} \approx 6 \times 10^{-12}$ which gives a reason to the question

But as we know that the accelerated charge particle radiate electromagnetic radiation . . . . . . Why we don't take it into account in our calculations?

  • $\begingroup$ That's great! Thanks for Help. $\endgroup$ – Young Kindaichi Feb 20 '20 at 8:11

Hope this link might help you.

Does a constantly accelerating charged particle emit EM radiation or not?

This relationship is not an absolute formula.It is just a precise relationship , which has been derived experimentally. Also it is not only frequency $\nu$ mentioned but its maximal value $\nu_{max}.$

This formula only tells you maximum frequency.which you will get corresponding to small energy losses.

In the experiment also when you are applying such a high voltage , and the kinetic energy is very much huge and its take negligible time to cross the tube in (very minimal time).so one can very much safely ignore that.

  • $\begingroup$ There is also the work function, which is usually ignored. That is a much larger effect than any radiation loss due to electron acceleration. $\endgroup$ – Pieter Feb 19 '20 at 17:45

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