Chapter Notes: X-Rays Physics Class 12



Notes for X-Rays chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

 

When a heavy metal target is bombarded with high-energy (30 – 50keV) electrons, it emits X rays. The radiation involves both a continuous and a line spectrum, as shown in the figure. The continuous spectrum, which starts at some minimum wavelength {\lambda _0}, arises from the rapid deceleration of the electrons when they enter the target – it is called bremsstrahlung, or braking radiation. The existence of minimum wavelength (or maximum frequency) is further evidence in favour of the photon concept. The highest frequency photon is emitted when an electron loses all its energy in one step. By equating the energy of the electron (eV) to the energy of the photon (h{\nu _0}), we find
h{\nu _0} = eV
or {\lambda _0} = {{hc} \over {eV}}

The minimum wavelength depends on the electron energy, but not on the target material.
The line spectrum depends on the element used as target. These characteristic X rays are produced when an electron knocks out an atomic electron from one of the inner levels. The ejected electron leaves a vacancy, which is then filled by an electron falling from a higher level. In the process a high  energy photon is emitted. If the transitions are to the n = 1 level, the X rays are labeled K\alpha , K\beta …….If they are to the n = 2 level, they are labeled L\alpha , L\beta , … etc. In the adjoining figure the energy level diagram for an atom is shown. The arrows indicate the transitions that give rise to the different series of X-rays.

In 1913, Moseley noted that the characteristic lines shifted systematically as the target material was changed. He plotted the square root of the frequency of the K\alpha line versus the atomic number Z for many elements. The straight line he obtained is shown in the figure.

Moseley’s plot did not pass through the origin. Let us see, why?. Once one of the two electrons in the n = 1 level is ejected, an electron in the next highest level will drop to the lower state to fill the vacancy and in the process it emits the K\alpha frequency. For this electron the electric field due to the nucleus is screened by the remaining electron in the n = 1 level. Moseley estimated that the effective nuclear charge for the K\alpha transition is (Z – 1)e. Thus Moseley’s law for the frequency of the K\alpha line is

\sqrt {{\nu _{{K_\alpha }}}} = a\left( {Z - 1} \right)
where a = \sqrt {{3 \over 4}Rc}    in which R is the Rydberg constant c is the speed of light.

The wavelength of K – lines is given by

{1 \over \lambda } = {\left( {Z - 1} \right)^2}\left[ {1 - {1 \over {{n^2}}}} \right]   where  n = 2, 3, 4,…………  

Example 1
Find the cut-off wavelength of the X-rays emitted by an X-ray tube operating at 30 kV.

Solution:

For minimum wavelength, the total kinetic energy should be converted into an X-ray photon.
Thus,

\lambda = {{hc} \over E} = {{12400} \over E} = {{12400} \over {30 \times {{10}^3}}} = 0.41 Å

Example 2
Show that the frequency of K\beta X-ray of a material equals to the sum of frequencies of K\alpha and L\alpha X-rays of the same material.

Solution:

The energy level diagram of an atom with one electron knocked out is shown above.
Energy of K\alpha X-ray is {E_{{K_\alpha }}} = {E_L} - {E_K}
           of K\beta  X-ray is {E_{{K_\beta }}} = {E_M} - {E_K}
and,     of L\alpha  X-ray is {E_{{L_\alpha }}} = {E_M} - {E_L}
thus,  {E_{{K_\beta }}} = {E_{{K_\alpha }}} + {E_{{L_\alpha }}}
or      {\nu _{{K_\beta }}} = {\nu _{{K_\alpha }}} + {\nu _{{L_\alpha }}}

 



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