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Expert-verified Found in: Page 1251 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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.

It is shown that the cutoff wavelength in the continuous x-ray spectrum from any target is given by ${\lambda }_{min}=1240/V$.

See the step by step solution

## Step 1: The given data

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

## Step 2: Understanding the concept of Plank’s relation:

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,

$\mathrm{h}=6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\phantom{\rule{0ex}{0ex}}=\left(6.242×{10}^{15}×6.63×{10}^{-34}\right)\mathrm{keV}.\mathrm{s}\phantom{\rule{0ex}{0ex}}=41.384×{10}^{19}\mathrm{keV}.\mathrm{s}$

The speed of light is,

$c=3×{10}^{8}m/s\phantom{\rule{0ex}{0ex}}=\left(3×{10}^{8}×{10}^{12}\right)\mathrm{pm}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=3×{10}^{20}\mathrm{pm}/\mathrm{s}$

The energy generated due to accelerating potential,

$\Delta E=eV$ ….. (2)

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

## Step 3: Calculation of the cut-off wavelength:

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:

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

${\lambda }_{min}=\frac{hc}{eV}\phantom{\rule{0ex}{0ex}}=\frac{\left(41.384×{10}^{-19}keV.s\right)\left(3×{10}^{20}pm/s\right)}{eV}\phantom{\rule{0ex}{0ex}}=\frac{1240keV.pm}{eV}\phantom{\rule{0ex}{0ex}}=\frac{1240pm}{V}$

Here, the potential V is in the kilovolts.

Hence, it is proved that the wavelength value is ${\lambda }_{min}=\frac{1240}{V}$. ### Want to see more solutions like these? 