# X-ray spectrum meaning help

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#1
Hi. Please could someone explain the X-ray spectrum please? I have attached an example picture from my textbook. This might be completely wrong, but my understanding is that on the x-axis is the wavelength of the emitted X-rays which changes when the supply voltage in the tube is altered (my guess is that a higher voltage results in a smaller wavelength/higher frequency of the X-ray...?). And as you go further down the x-axis the wavelength of the X-ray increases, and so the frequency of the X-ray decreases. This means means the X-ray has less energy (E = hf) and therefore has a stronger intensity. The spikes (k-lines) and caused because at that point (amount of voltage...?) the bombarding electrons remove electrons atoms very close to the nucleus. So the gaps are quickly filled up by electrons on higher energy level going to lower energy levels. This burst in energy is emitted of X-ray of photons of specific wavelengths and so there is a burst in the intensity at this point.
All of this was just what I think happens and I expect it to be largely wrong. So please correct me if you can. Thanks.
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4 years ago
#2
First, ignoring the spikes, the graph is a distribution of wavelengths (energies) with a sharp cut off at short wavelength (high energy)

Remember how x-rays are formed - they are released when charged particles undergo steep accelerations or deceleration. X-ray machines fire high speed electrons at a metal target. When they hit the target they decelerate quickly and release some of their energy as X-rays (some also to heating of the metal).

The electrons don't all undergo the same deceleration when striking the target, which accounts for the distribution. The sharp cut off represents the highest energy x-rays produced when electrons lose all of their energy in one collision. Some only lose part of it, and hence the xrays have a range of energies and wavelengths.,

Second, the spikes. They correspond to energy levels - the emissionspectrum - of the metal of which the target is made, in your case molybdenum. The incident electron knocks an electron out of its position in the metal atom and this electron is then quickly replaced by others which results in emission of energy and hence the spike.

Your comment about the voltage is correct - the acccelerating voltage determines the kinetic energy which the electrons have before striking the target and therefore determines where that cut off wavelength is.
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#3
(Original post by phys981)
First, ignoring the spikes, the graph is a distribution of wavelengths (energies) with a sharp cut off at short wavelength (high energy)

Remember how x-rays are formed - they are released when charged particles undergo steep accelerations or deceleration. X-ray machines fire high speed electrons at a metal target. When they hit the target they decelerate quickly and release some of their energy as X-rays (some also to heating of the metal).

The electrons don't all undergo the same deceleration when striking the target, which accounts for the distribution. The sharp cut off represents the highest energy x-rays produced when electrons lose all of their energy in one collision. Some only lose part of it, and hence the xrays have a range of energies and wavelengths.,

Second, the spikes. They correspond to energy levels - the emissionspectrum - of the metal of which the target is made, in your case molybdenum. The incident electron knocks an electron out of its position in the metal atom and this electron is then quickly replaced by others which results in emission of energy and hence the spike.

Your comment about the voltage is correct - the acccelerating voltage determines the kinetic energy which the electrons have before striking the target and therefore determines where that cut off wavelength is.
Thanks a lot for explaining, it was really helpful and I now understand the spectrum far better. But the remaining issue that I have is why does the intensity not increase as the frequency and energy of the X-ray increases? Also the x-axis can be labelled with Photon energy instead of wavelength couldn't it, like this?

Edit: Is it because those specific wavelengths correspond to where the electrons collide in the atom for that specific material? And the rise in intensity is attributed to the number of photons released, rather to the frequency of the photons released?
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4 years ago
#4
I think it's better to think of it like that, yes - the higher the intensity the more rays of that wavelength are produced - or perhaps if you prefer the total intensity of all the photons of that particular wavelength.
Remember that this shape of curve is typical of a statistical distribution.
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#5
(Original post by phys981)
I think it's better to think of it like that, yes - the higher the intensity the more rays of that wavelength are produced - or perhaps if you prefer the total intensity of all the photons of that particular wavelength.
Remember that this shape of curve is typical of a statistical distribution.
Thanks! I was really confused because I originally thought it meant those individual photons had that intensity, not that there were more photons at that specific wavelength. Which really confused my because the specific photons which have a higher frequency have more energy and therefore a higher intensity. Anyway it makes sense now. Thankyou!
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