X-Ray Wavelength and Frequency

x-rays are produced in an x-ray tube where high velocity electrons are impinged on a

target, the electrons have inelastic collisions with the target nuclei causing the electrons to slow down emitting electromagnetic radiation, inelastic collisions with orbital electrons cause ionization & characteristic x-rays. the atomic number, Z, of the target and the participating electron shell ie: K or L will determine the emission energies.

rule of thumb> E = (13.6 eV) time Z squared times (1/n1 sqd - 1/n2 sqd)

you wish to calculate it by formula: E = h . v ?

Its not answered by s simple words. it has not only concerned with high voltage, but with target material. and transfer coefficiency, which is very lower about 1.5%.

as #2 said the atomic number, Z, of the target and the participating electron shell ie: K or L will determine the emission energies.

E = 13.6 eV * Z^2 * (1 - 1/2^2) = 10.8 * Z^2

is only mean how myuch enegy can produce. for an element

different voltage will hit defferent shell like k and l. and have different intensity.

generally we say in term of kv/A, such like 100Kv/20mA energy set or units etc.(here the word energy means power in practice, but people in the field used to call it as this. hehe, dont mistake) only accelerator use MeV as unit.

of cause once you apply 100kv across on the tube, the electron has 100keV energy in this firld, but x-ray intensity emission will be .another number.

you'd better refer to atomic physics for details. for example

In a common X ray tube, applying "V" volts DC will produce 0% "V" eV Xrays, and linearly rising componenets of Xrays of lesser energy. The average energy of the emitted spectrum is about 1/3 of the voltage applied (viz Ave energy = "V"/3 eV

"V" is in the order of 30-150,000 volts.

For any particular voltage "V" the lowest energy emitted xray will be zero eV, and for each collision that does not produce "V" electron volt Xrays, second and third ... collisions can convert the rest of the incident enegy to lesser energetic components, but most of the collision energy is given off as heat. Hence the annode of Xray tubes are cooled by forced convection of water, air or oil.

The very low energy xrays {up to10KeV} hardly penetrate the Xray tube. Ony a few % of the total energy (V x I x t)are converted to useful Xrays. There will be enhanced emission in the way of a peak at the voltages of the K,L and possibly M binding energies of the target material eg characteristic peak of tungsten at about 60 keV

Hi all,

An X-ray tube produces a rather continuous spectrum of wavelengths. A great fraction of energy is produced as heat (That's the reason X-ray tubes have a Wolfram/Tungsten target and even thus, sometimes need cooling media)

The voltage applied determines the maximum speed of electrons which hit the target and therefore the maximum energy of X-ray achievable (When all energy of electron is lost in just one collision and transformed integrally in radiation)

The minimum wavelength you can get is given by Duane Hunt equation:

λ_min =12.34/V, where λ is wavelength in Angstrom and V the tube voltage in kV

But you will get few density of photons with such a maximum energy. The maximum density of monochromatic photons will have a wavelength about twice the minimum calculated by Duane Hunt equation.

Sorry if I give you just the wavelengths, but I'm lazy and I don't remember just now the Planck's constant to make the conversion to energy.

Kind regards

Hey Kwetz,

Thanks! That was informative!

the large energy it has, the depth it can penetrate. this energy depends on its wave length, or say photon energy, the short wave length, the higher it has energy. not on its indtensity.

this energy can be get by increase voltage, but intensity can be get by increas current.

the higher the voltage, we call it the more hardness the ray has.

you can get from formula, I listed above.

E= h. v, where v (neu) = frequency, h pkcontant number and E, energy

you can write in eU, where u tube votlage Kv, e, electron electric charge

from this you can estimate the smallest wave lenght or themax frequency number.

1/3 inch steel plate, can be penetrated by 80--100kv voltage.

but if you hope to display on photograph. you need more intensity.

the large energy it has, the depth it can penetrate. this energy depends on its wave length, or say photon energy, the short wave length, the higher it has energy. not on its indtensity.

Thank you, I knew it

What I was trying to say is that from the "almost" continuous (quantum theories) spectrum, the maximum energy of X-ray (the minimum wavelength) is voltage dependent. If you increase the tube current while keeping voltage constant, the spectrum in wavelength (photon energy) is the same, but you get more intensity of each and every wavelength.

If you increase the voltage, the kinetic energy of electrons is higher, so they can transfer in just one interaction more energy and the maximum possible X-ray energy increases as well as you get more intensity of each and every wavelength.

The two classic figures illustrated it too well:

Obviously, the maximum theoretical energy is a threshold (intensity or photon nº is near zero) and the maximum intensity of radiation → the greater number of photons of the same wavelength occurs at about twice the minimum wavelength.

I was talking not just of individual photons but also of global behaviour of X-ray spectra.

Kind regards

you are not wrong. what I said is aim at #7 issue, which is very easily answered by any my staffs and workers (but I almost forget all). its relationship among voltage, current, hardness and intensity etc.

The limit wavelength you deduced is right λ_min =12.34/V, if in nm will be = 1.2(42).(tolerant may come from values of light speed value,e and h. it doesnt matter)

but I really dont know who invented the formula until you metioned its Duane Hunt equation

we usually deduce it by energy equation.

Hi cnpower,

Yes, theis will provide a photo image on the other side of the steel.

hm? not bad, you know this important principle

X-rays are usually generated by slamming an electron beam into a dense metal which causes a fast deceleration of electron beam and heating of the target. This action causes a broad spectrum of energy to be generated (Bremmstralung effect).

The electron beam voltage is measured in peak kV (kVp). To generate 100 keV x-rays the electron beam must be a minimum of 100 kVp. The energy of the x-rays will extend from a threshold of around 5 keV up to the accelerating limit of 100 keV. The resulting energy curve rises fast at the low end, reaching a maximum at 1/3 to 2/3 of the accelerating potential, and then tapers to the maximum potential. The bulk of the x-rays energy will be located in the middle half of the curve. Depending on the target material there will be one of two prominent peaks in the spectrum, characteristic of the target material.

For a 100kVp electron beam and a tungsten target: the bulk of the energies will be located around 30 keV to 60 keV with two very prominant peaks around 57 keV and 69 keV. The maximum x-ray energy will be 100 keV.

Welcome aboard! I voted you a GA (good answer) for that first post because from what I can tell, it was one. Don't expect each and every post to get a GA vote, though...

hehe, its too unfaire, maybe too cups of beer. Kwetz's answer is also a GA. very clear, why not get a vote? I vote him as well.