# X-Ray Diffraction

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?

• That's great! Thanks for Help. – Himanshu Sahu Feb 20 at 8:11

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 $$\nu_{max}.$$