Question

If 60 kV potential difference is applied on Coolidge tube, then calculate minimum wavelength of continuous X-ray spectrum.

Answer 1

Energy of an electromagnetic radiation is given by:
$E = h\nu = \dfrac{{hc}}{\lambda }$ (by Planck’s law)
Where, h is Planck’s constant and $h = 6.63 \times {10^{ - 34}}{\text{ J - s}}$.
C is the speed of light and $c = 3 \times {10^8}{\text{ m}}{{\text{s}}^{ - 1}}$.
And $\lambda$ is the wavelength of the electromagnetic radiation.
Also, we know the energy of an accelerated charge particle is given by:
$E{\text{ }} = {\text{ }}qV$
Where, q is the charge of the particle and V is the potential through which it has been accelerated.
Here, in the given question $E{\text{ }} = {\text{ }}eV$(in Joules or eV), since the charge particle is an electron.
Equate these formulas for E and find the minimum wavelength of X-ray spectrum.

Complete step by step answer:
As we know, energy of an electromagnetic radiation is given by:
$E = h\nu = \dfrac{{hc}}{\lambda }$ (by Planck’s law) ------------> (i)
Where, h is Planck’s constant and $h = 6.63 \times {10^{ - 34}}{\text{ J - s}}$.
C is the speed of light and $c = 3 \times {10^8}{\text{ m}}{{\text{s}}^{ - 1}}$.
And $\lambda$ is the wavelength of the electromagnetic radiation.
Since, the maximum energy of the x-rays is the same as the kinetic energy of the electrons.
Therefore, the energy of the accelerated charge particles is given by:
E = qV
Here, E = eV (in Joules or eV), since the charge particle is an electron. ------------->(ii)
From equating eq.(i) and (ii) we get:
$\begin{gathered} \dfrac{{hc}}{\lambda } = eV \\\ \Rightarrow {\lambda _{\min }} = \dfrac{{hc}}{{e{V_{\max }}}}{\text{ }}\left( {\because {\text{ the energy is maximum, }}\lambda {\text{ will be minimum}}} \right) \\\ \end{gathered}$
Here, potential difference by which the electron accelerated is= 60 kV= 60000 V
So, the minimum wavelength of X-ray spectrum is:
${\lambda _{\min }} = \dfrac{{hc}}{{e \times 60000}}$
Putting value of,
h = 6.63 x 10-34 J-s
e = 1.6 x 10-16 C
and c =3 x 108 ms-1 we get:
${\lambda _{\min }} = \dfrac{{12420}}{{60000}}\dot {\rm A} = 0.21\dot {\rm A}$

Note:
Working of Coolidge tube:
When the cathode filament of Coolidge tube is heated, it emits electrons. And as hotter the filament gets, the greater the cathode emits electrons. These electrons being negatively charged are accelerated towards the positively charged anode and when the electrons strike the anode, they change their direction and emit x-rays with a continuous range of energies. And the maximum possible energy of the x-rays is the same as the kinetic energy of the electrons striking the anode.