Question

In a Coolidge tube, the potential difference used to accelerate the electrons is increased from \[12.4KV\] To \[24.8KV\] . As a result, the difference between the wavelengths of the ${K_\alpha }$ line and the minimum wavelength increases three times. The wavelength of ${K_\alpha }$ the line is \[\left( {\dfrac{{hc}}{e} = 12.4KV{A^0}} \right)\]
A. $1{A^\circ }$
B. $0.5{A^\circ }$
C. $1.5{A^\circ }$
D. \[1.25{A^\circ }\]

Answer 1

Hint: Here initially we will study about Coolidge tube and based on the selection of cathode and anti-cathode material, and then solve the above problem using the formula that is given in the question and equating it to thrice the amount. Hence we can get the required solution.

Complete step by step solution:
The wavelength of the x-rays discharged from a Coolidge tube is generally found by the material its anode is made up of. X-rays are created in a Coolidge tube. In the tube, a cathode (a plate linked to the negative terminal of a high voltage power supply) is heated by a filament by which DC is passed through a small dc voltage source to provide thermionic elections. The high potential difference V retained between the cathode and a metallic target accelerates the electrons in the direction of the latter. The target’s face is at an angle of \[60^\circ \] to the direction of the electrons. The tube is emptied to ${10^{^ - 4}}mm$ of mercury or less pressure.
The X-rays formed by the tube are of two types, continuous and characteristic. Though the previous depends on the accelerating voltage \[V\], the final depends on the target user.
This minimum wave-length resembles the maximum energy of the X-rays which is the same as the maximum kinetic energy eV of the bombarding electrons. The kinetic energy of these electrons at an accelerating potential difference of \[V{\text{ }}volts{\text{ }} = eV\]
When all this energy is transformed into X-rays, the frequency\[\;f\] is directed by the relation\[ = {\text{ }}h \times f\] here\[\;h\] is the Planck Constant.
Therefore,
\[hf = eV\]
Where \[\dfrac{{\left( {hc} \right)}}{\lambda }{\text{ }} = {\text{ }}eV\]
So according to the question
\[3\left( {{\lambda _{{K_\alpha }}} - \dfrac{{12.4}}{{12.4}}} \right) = {\lambda _{{K_\alpha }}} - \dfrac{{12.4}}{{12.4}}\]
\[ \Rightarrow {\lambda _{{K_\alpha }}} = 1.25{A^ \circ }\] Hence option D is correct

Note:
Electromagnetic theory calculates that an accelerated charge emits electromagnetic waves, and fast-moving electrons when brought to abrupt rest is violently decelerated. X-rays formed are called Bremsstrahlung or braking radiation. These radiations are more significant for electrons than heavier charged particles since electrons are more violently decelerated when it moves near nuclei in their route. The nonstop X-rays produced for accelerating voltage \[V\]differ in wavelength, but nothing has a wavelength shorter than a definite value ${\lambda _{min}}$.