# Intensity - wavelength graph for X-ray

This is an intensisity vs. wavelength graph of X-ray emmision.

The cut-off wavelength is the minimum wavelength of the X-ray emitted. But doesn't minimum wavelength correspond to maximum energy? Why is intensity zero at minimum wavelength? Shouldn't it be maximum when wavelength is minimum since $\lambda=\frac{hc}{E}$

• Don't understand your reasoning. Why should the intensity be a maximum at minimum wavelength because of the equation $\lambda$=hc/E? – user93237 Feb 1 '16 at 6:53
• @SamuelWeir Intensity is energy per unit area per unit time. So isn't intensity proportional to energy? – Aditya Dev Feb 1 '16 at 7:17

You are confusing the energy of a single photon with the total energy (intensity) in a differential wavelength band $\lambda$ to $\lambda + \epsilon$ . Yes, the shortest wavelength corresponds to the photon with greatest energy, but the X-ray machine generates far more photons at longer wavelengths. Thus the peak intensity occurs where the product of photon energy and number of photons is maximized.

• That explains why It doesn't correspond to a maxima at the cutoff wavelength. However, why is the intensity zero? – satan 29 Jul 23 '20 at 19:03

Your graph is a standard one to show the spectrum of wavelengths emitted from an X-ray tube.
The X-rays are produced by getting energetic electrons hit a metal target.
The electrons are first accelerated by being attracted to a positive anode which is at a high potential $V$ relative to the negative cathode from which they are emitted.
The kinetic energy of these electrons is $eV$ where $e$ is the charge on the electron.
When the high energy electrons hit the metal target on the anode they are slowed down very rapidly and in doing so emit electromagnetic radiation (photons). In general not all of the electron's kinetic energy $eV$ is converted into a single photon. However if all all the kinetic energy of one electron was converted into one single X-ray photon this would represent the maximum energy and hence maximum frequency $f_{\text{max}}$ (or minimum wavelength $\lambda_{\text{min}}$) that an X-ray photon could have.
$eV = h f_{\text{max}}=\frac {hc}{\lambda_{\text{min}}}$

Photons having more energy than this cannot be produced as the probability of two electrons giving up their kinetic energy to produce one photon is very. very small.

• True but doesn't answer the question. – Carl Witthoft Feb 1 '16 at 13:59

I would like start answering this question by explaining Intensity here .Because ,In previous answer,people seemed to get misunderstood by Intensity of X-ray with **Intensity of charged electron ** (energy due to accelerated electron ).Now ,Supposing electron from cathode are accelerated by Applying potential differences “V” ,so total energy of one electron will be eV .Now ,This electron hits the target and loose the energy, amount of loosed energy will convert into photon of corresponding wavelength (Energy =hc/λ).Remember ,Maximum energy electron can loose is eV ,so minimum correspond wavelength (hc/λ min =eV ,So λmein= hc /eV ). There will be no photon of having wavelength (λ min) or in other words,There is zero probability of this wavelength of X-ray .That is why ,Intensity at λ min corresponds to zero . I hope ,This answer will help you.