Question

An X-ray tube is operated at $6.6KV$. In the continuous spectrum of the emitted X-rays, which of the following frequencies will be missing?
$\text{A}\text{. }{{10}^{18}}Hz$
$\text{B}\text{. }1.5\times {{10}^{18}}Hz$
$\text{C}\text{. }2\times {{10}^{18}}Hz$
$\text{D}\text{. }2.5\times {{10}^{18}}Hz$

Answer 1

Hint: For finding the missing frequencies, we need to find the value of maximum frequency obtained in the continuous spectrum of the emitted X-rays. The frequencies above the maximum frequency will be missing from the spectrum.

Formula used:

${{\upsilon }_{\max }}=\dfrac{eV}{h}$

Complete step by step answer:
A spectrum having no apparent breaks or gaps throughout its wavelength range is called a continuous spectrum. The continuous X-ray spectrum is defined as the range of photon energies produced in an X-ray tube as a result of the properties of Bremsstrahlung radiation. The energy of X-ray photons can take a value from zero to the maximum kinetic energy of the incident electrons. Continuous X-ray spectrums are created when a free moving electron gets in close enough range of an atomic nucleus that the two electromagnetically interact. When this occurs, the electron loses some of its kinetic energy.

If electrons are accelerated to a velocity $v$ by a potential difference $V$ and then allowed to collide with a metal target, the maximum frequency of the X-rays is given by:

$m{{v}^{2}}=eV=h\upsilon$

Therefore, X-ray frequency $\upsilon =\dfrac{eV}{h}$

Where,
$e$ is the charge on an electron
$V$ is the accelerated voltage
$h$ is the Planck's constant

Maximum frequency is directly proportional to the accelerating voltage.
In the above question,

Voltage is given as $V=6.6KV=6600V$
Therefore, ${{\upsilon }_{\max }}=\dfrac{eV}{h}$

Putting values in above equation,

$\begin{aligned} & {{\upsilon }_{\max }}=\dfrac{1.6\times {{10}^{-19}}\times 6600}{6.6\times {{10}^{-34}}} \\\ & {{\upsilon }_{\max }}=1.6\times {{10}^{18}}Hz \\\ \end{aligned}$

Frequencies above ${{\upsilon }_{\max }}$ will be missing from the continuous spectrum of the emitted X-rays.

Hence, correct options are C and D.

Note: The higher frequency in the emission rays corresponds to a value of higher energy, resulting in a shorter wavelength. As we calculated the value of maximum frequency, we can calculate the value of minimum wavelength associated with a continuous spectrum of X-rays in an accelerated voltage.