Question

Find the cutoff wavelength for the continuous X-rays coming from X-ray tubes operating at 40 Kv.

Answer 1

Hint: Wavelength is mathematically defined as the ratio of Velocity to the frequency. In theoretical terms wavelength means the distance between the identical points of the wave or we can say that the wavelength is the spatial period after which the wave starts repeating itself after a particular interval of time. An electromagnetic wave can be distinguished in the terms of frequency and wavelength.

Complete answer:
We have been given that V = 40Kv
In the above question we need to find the cut off wavelength of X-Ray for a given energy of X-ray tube.
We know that the formula of wavelength can be given as $\lambda =\dfrac{hc}{E}$ ……… (1)
Where “E” is the Energy
“C” is the speed of light
h = Planck's Constant
Which have a value of \[4.1357\times {{10}^{-}}^{15}~eV\text{ }s\]
Energy Can be defined as E = eV
Where “e” is the charge of electron which is equal to \[1.602176634\text{ }\times \text{ }{{10}^{-}}^{19}C\]
Putting the value of E in Equation (1) We get,
$\lambda =\dfrac{hc}{eV}$
Putting the Values in above equation we get,
$\lambda =\dfrac{(4.1357\times {{10}^{-}}^{15}~eV\text{ }s)(3\times {{10}^{8}}m{{s}^{-1}})}{(1.602176634\text{ }\times \text{ }{{10}^{-}}^{19}C)(40Kv)}$
$\lambda =311\times {{10}^{-4}}nm$
This is the cutoff wavelength for the X-rays coming from X-ray tubes operating at 40 Kv.
Hence, we can conclude that the correct answer of the above question will be $\lambda =311\times {{10}^{-4}}nm$.

Note:
The terms cut off wavelength is basically used in optical fiber communication. Cut off wavelength is the maximum wavelength at which the fiber optics will operate in a single mode fiber. We can precisely say that cut off wavelength is related to single mode fiber communication. Optical fibers work on the principle of total internal reflection and are the most useful product in our life.