Question

An electron beam in X-Ray tube is accelerated through a potential difference of $50000\,Volt$. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is
A) $2.5\,{A^0}$
B) $0.25\,nm$
C) $0.25\,cm$
D) $0.025\,nm$

Answer 1

Hint: The shortest wavelength of the X-ray will have the maximum frequency. Since energy is directly related to frequency, we need to calculate the wavelength of the X-ray with maximum energy. X ray with maximum energy will be emitted when an electron with maximum kinetic energy strikes the target. Hence, by calculating the maximum kinetic energy of the electron and equating it with the energy of the X-ray we can find the answer to this question.

Complete step by step solution:
It is given that an electron beam is accelerated using a potential difference of $50000\,volt$.
Thus,
$V = 50000\,volt$
When the electron beam is allowed to fall on a tungsten target x-ray will be emitted. We need to find the shortest wavelength of the X-ray emitted by the tube.
We know that wavelength is inversely related to frequency. So, minimum wavelength implies maximum frequency.
Since energy of radiation is given as
$E = h\upsilon $
Where h is the Planck's constant and $\upsilon $ is the frequency.
$E = h\dfrac{c}{\lambda }$
Since, $c = \upsilon \lambda $
Where c is the velocity of light $\upsilon $ is the frequency and $\lambda $ is the wavelength.
Since frequency is directly related to the energy, the maximum frequency will correspond to the X-ray with maximum energy.
We know that the kinetic energy of the incoming electron is transferred as the energy of the outgoing X-ray.
So, if you find the maximum kinetic energy of the incoming electron, we can use it to find the answer to this question.
The kinetic energy of the electron is supplied by the potential difference.
Energy due to application of potential difference is given as the product of charge $q$ and the potential difference $V$ .
 $ \Rightarrow U = qV$
 $ \Rightarrow U = eV$
Where, $e$ is the charge of electrons and $V$ is the potential difference.
Hence kinetic energy will be equal to this value.
$ \Rightarrow KE = eV$
On substituting the value of potential, we get
$ \Rightarrow KE = e \times 50000$
This will be the same as the energy of the X ray with the shortest wavelength.
Thus, we can equate both these values.
$ \Rightarrow h\dfrac{c}{\lambda } = eV$
$ \Rightarrow h\dfrac{c}{\lambda } = 50000\,eV$
In terms of electron volt, the energy of photon is given as
$E = \dfrac{{12400\,eV{A^0}}}{\lambda }$
On substituting all values, we get
$ \Rightarrow \dfrac{{12400\,eV{A^0}}}{\lambda } = 50000\,eV$
$ \Rightarrow \lambda = \dfrac{{12400\,eV{A^0}}}{{50000\,eV}}$
$ \Rightarrow \lambda = 0.248\,{A^0}$
$ \Rightarrow \lambda \approx 0.25\,{A^0}$
We know that \[1\,{A^0} = {10^{ - 10}}\,m\] and $1\,nm = {10^{ - 9}}m$
$ \Rightarrow \lambda \approx 0.025\, \times {10^{ - 9}}\,m$
$ \Rightarrow \lambda \approx 0.025\,\,nm$
This is the shortest wavelength of the X-ray emitted.

Hence the correct answer is option (D).

Note: Remember that the wavelength and frequency are inversely related. Hence the energy of the shortest wavelength will be the maximum, since energy is directly related to frequency. That is why we calculated the maximum kinetic energy that is transferred by the electron. This maximum kinetic energy will be the maximum energy of the X-ray emitted.