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What will be the wavelength of an X-ray, if we apply a potential difference of, say $\pu{20 kV}$ across the ends of the X-ray tube?

Alright, I started with the modified de-Broglie equation- $$\Lambda=\frac{h}{\sqrt{meV}}$$ I pinned the values of Planck constant, electronic charge and potential difference but can't guess what would be the mass $m$ here.

The answer comes $\pu{0.62*10^-10m}$ if solved perfectly. Can anybody confirm?

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    $\begingroup$ check Planck–Einstein equation. $\endgroup$ – permeakra Jul 14 '14 at 8:34
  • $\begingroup$ But actually, the question provides insufficient information. X-ray from X-ray tubes is part of hi-energy electronic spectra of the anode material, so in reality the wavelength of most photons would be defined by anode material, and the only requirement for the potential difference is to be enough to cause excitation of the anode material. (It is somewhat oversimplified description, but reasonably adequate) $\endgroup$ – permeakra Jul 14 '14 at 8:39
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Firstly as @MaxW pointed out, using the given information, it is possible to find the shortest wavelength (or maximum frequency) X-ray only.

Spectrum of the X-rays emitted by an X-ray tube with a rhodium target, operated at 60 kV. The smooth, continuous curve is due to bremsstrahlung, and the spikes are characteristic K lines for rhodium atoms.

In an X-ray tube, electrons are accelerated in a vacuum by an electric field and shot into a piece of heavy metal (e.g., $\ce{W, Rh, Mo, Cu, Ag}$) plate. X-rays are emitted as the electrons decelerate in the metal. The output spectrum consists of a continuous spectrum of X-rays, with sharp peaks at certain energies as in the graph. The continuous spectrum is due to bremsstrahlung (German for "deceleration radiation"), while the sharp peaks are characteristic X-rays associated with the atoms in the target.

The spectrum has a sharp cutoff at low wavelength (high frequency), which is due to the limited energy of the incoming electrons (which is equal to the voltage on the tube times the electron charge). This cutoff applies to both the continuous (bremsstrahlung) spectrum and the characteristic sharp peaks, i.e. there is no X-ray of any kind beyond the cutoff. However, the cutoff is most obvious for the continuous spectrum.

Hence, the Duane-Hunt law comes handy in here, which gives the maximum frequency of X-rays that can be emitted by Bremsstrahlung in an X-ray tube by accelerating electrons through an excitation voltage $V$ into a metal target.

The maximum frequency $\nu_\text{max}$ is given by-

${\displaystyle \nu _{\text {max}}={\frac {eV}{h}},}$ which corresponds to a minimum wavelength

${\displaystyle \lambda _{\text {min}}={\frac {hc}{eV}},}$ where $h$ is Planck's constant, $e$ is the charge of the electron, and $c$ is the speed of light.

Thus putting the values given-

$$\lambda _{\text {min}}={\frac {hc}{eV}}={\frac {\pu{1240 eV nm}}{\pu{20*1000 eV}}}= \pu{0.62*10^-10 m}$$

For further references check the Wikipedia page.

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  • $\begingroup$ I speculate what reads like a (of course incomplete) list of (solid) targets required the correction Mb -> Mo, and benefits the addition of Cu and Ag so often used for the characterization of organic molecules (proteins), regardless if «classical» tube, rotating anode (especially in biochem), or microfocus X-ray tube. $\endgroup$ – Buttonwood Sep 8 '20 at 20:58
  • $\begingroup$ @Buttonwood Ah...sure! I was in a hurry, and I simply omitted that part while doing the corrections. Thanks! I wasn't aware of the usage of $\ce{Ag}$ though. $\endgroup$ – Sir Arthur7 Sep 9 '20 at 3:22
  • $\begingroup$ Some of them are listed in IUCr's teaching, and of course by the manufactures of X-ray tubes. Beside micro focus tubes (e.g. on diffractometers with both a Cu and a Mo source), be aware of the brighter liquid targets, e.g. background for small crystals. $\endgroup$ – Buttonwood Sep 10 '20 at 20:39
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$$\lambda=\frac h{\sqrt{meV}}$$ $$E=mc^2$$ Also $$E=\frac{hc}\lambda$$ Therefore, $$mc^2=\frac{hc}\lambda$$ $$mc=\frac h\lambda$$ Hence $$m=\frac h{\lambda c}$$ Plugging the value of $m$ from here we get an equation with all known constants. Solving for $\lambda$, we can get the result as $0.62\times10^{-10}\ \mathrm m$.

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  • $\begingroup$ This gives the shortest (highest energy) x-ray. The overall spectra from the x-ray tube would be Bremsstrahlung radiation (sort of like blackbody radiation) with line spectra from the element of the anode material of the x-ray tube. The electron beam of an x-ray tube is in a vacuum. The x-rays are generally emitted through a thin beryllium window which absorbs lower energy (long wavelength) x-rays. $\endgroup$ – MaxW Oct 22 '15 at 1:27

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