Show that the cutoff wavelength in the continuous x-ray spectrum from any target is given by ${\mathbf{\lambda }}_{\mathbf{min}}\mathbf{=}\mathbf{1240}\mathbf{/}\mathbf{V}$, where is the potential difference (in kilovolts) through which the electrons are accelerated before they strike the target.

A continuous x-ray spectrum from any target is produced due to the striking of the electrons that are accelerating before the strike.

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using the energy relation of Planck's equation and the energy difference created by the accelerating electron due to generated potential difference, to get the required equation of the cutoff wavelength in the continuous x-ray spectrum from any target.

Formulae:

The energy of the photon due to Planck’s relation,

$\Delta E=\frac{hc}{\lambda }$ ….. (1)

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} c=3\times {10}^{8}m/s \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3\times {10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

The energy generated due to accelerating potential,

$\Delta E=eV$ ….. (2)

Here,e is the charge and V is the potential.

As the accelerating electrons strike the target, they generate the same energy difference on the target as due to the accelerating potential. Thus, using equations (1) and (2), the cutoff wavelength in the continuous x-ray spectrum from any target is given by:

$eV=\frac{hc}{{\lambda }_{min}}$

$$\begin{gathered} {\lambda }_{min}=\frac{hc}{eV} \\ =\frac{\left(41.384\times {10}^{-19}keV.s\right)\left(3\times {10}^{20}pm/s\right)}{eV} \\ =\frac{1240keV.pm}{eV} \\ =\frac{1240pm}{V} \end{gathered}$$

Here, the potential V is in the kilovolts.

Hence, it is proved that the wavelength value is ${\lambda }_{min}=\frac{1240}{V}$.