Question

If the short wavelength limit of the continuous spectrum coming out of a Coolidge tube is 10Å, then the de Broglie wavelength of the electrons reaching the target metal in the Coolidge tube is approximately:
A. 0.3Å
B. 3Å
C. 30Å
D. 10Å

Answer 1

Hint: The maximum allowed energy of particles coming out of the Coolidge tube is fixed. The electron energy is converted into X-Rays. As a result, you can apply the energy conservation theorem to find the momentum of the electron. You can use the de Broglie hypothesis to find the wavelength of the electrons.

Formula Used:
The de Broglie wavelength is given by,
${{\lambda }_{de-Broglie}}=\dfrac{h}{p}$......................(1)
The kinetic energy of a particle is given by,
$KE=\dfrac{{{p}^{2}}}{2m}$.....................(2)
The energy of a photon is given by,
$E=\dfrac{hc}{\lambda }$....................(3)

Where,
${{\lambda }_{de-Broglie}}$ is the wavelength of the particle
$\lambda $ is the wavelength of the photon
$h$ is the Planck’s constant
$c$ is the speed of light
$p$ is the momentum of the particles
$m$ is the mass of the particle

Complete step by step answer:

The short wavelength limit of the continuous spectrum is given; hence the energy of the photon is,
$\dfrac{hc}{{{\lambda }_{\min }}}$

It should be equal to the Kinetic energy of the electrons,
$KE=\dfrac{{{p}^{2}}}{2m}$
Hence, we can write,
$\dfrac{hc}{{{\lambda }_{\min }}}=\dfrac{{{p}^{2}}}{2m}$
$\Rightarrow p=\sqrt{\dfrac{2mhc}{{{\lambda }_{\min }}}}$

So, the momentum of the electron is,
$p=\sqrt{\dfrac{2mhc}{{{\lambda }_{\min }}}}$

Now, we can find the de Broglie wavelength using the equation (1),
${{\lambda }_{de-Broglie}}=\dfrac{h}{p}$
$\Rightarrow {{\lambda }_{de-Broglie}}=h\times \dfrac{1}{\sqrt{\dfrac{2mhc}{{{\lambda }_{\min }}}}}$
$\Rightarrow {{\lambda }_{de-Broglie}}=\sqrt{\dfrac{h{{\lambda }_{\min }}}{2mc}}$

We know the following,
$c=3\times {{10}^{8}}$m/s
$h=6.626\times {{10}^{-34}}$ Js
${{\lambda }_{\min }}=10$ Å
$m=9.19\times {{10}^{-31}}$ kg

Hence, the de Broglie wavelength is,
$\Rightarrow {{\lambda }_{de-Broglie}}=\sqrt{\dfrac{(6.626\times {{10}^{-34}})(10\times {{10}^{-10}})}{2(9.19\times {{10}^{-31}})(3\times {{10}^{8}})}}$
$\Rightarrow {{\lambda }_{de-Broglie}}=0.3$ Å

So, the correct answer is - (A).

Note: The working principle of the Coolidge tube is as follows:
A heated filament emits electrons in the vacuum tube, which we call the Coolidge tube. The electrons are accelerated with the help of an electric field between the filament and the target. These electrons are absorbed in the target and it emits X-Ray. The energy of the electrons are transferred to the X-Ray. That is why we compared the kinetic energy with the energy of the photons.