Question

For modern X-ray tube, the wavelength of emitted X-rays and applied potential V of the tube are related as
A. $\lambda =\dfrac{12400}{V}A{}^\circ$
B. $\lambda =\dfrac{12400}{\sqrt{V}}A{}^\circ$
C. $\lambda =\dfrac{1240}{\sqrt{V}}A{}^\circ$
D. $\lambda =12400A{}^\circ$

Answer 1

You could simply recall the relation between the wavelength of this continuous X-ray spectrum and the accelerating potential containing other known constant values. Then, all you have to do is substitute these known values. Thereby, you will get the required relation in the form of one the given options.

Formula used:
$\lambda =\dfrac{hc}{eV}$

Complete step by step solution:
In the question, we are discussing the modern X-ray tube. We are supposed to find the relation between the wavelength$\lambda$ of the emitted X-rays and applied potential V of the tube. We know that an electron that is decelerated by Coulomb’s interaction of target atom emits X-rays in the continuous spectrum and the wavelength of this continuous X-ray spectrum is known to be related to the accelerating potential V as,
$\lambda =\dfrac{hc}{eV}$
Where, h is the Planck’s constant given by,
$h=6.63\times {{10}^{-34}}{{m}^{2}}kg/s$
c is the speed of light in vacuum given by,
$c=3\times {{10}^{8}}m{{s}^{-1}}$
e is the charge of an electron given by,
$e=1.6\times {{10}^{-19}}C$
Substituting all these values we get,
$\lambda =\dfrac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}V}$
$\therefore \lambda \approx \dfrac{12400}{V}A{}^\circ$
Hence, option A is found to be the answer.

Note: From the above acquired relation, it is very clear that the wavelength is inversely related to the accelerating potential. That is, when the wavelength increases then the accelerating potential increases. Now the intensity is known to be the rate of energy flow per unit area, also, the energy of photons is known to increase with increasing V. Therefore, intensity also increases with increase in applied potential.