Question

The potential difference applied to an X-ray tube is$V$. The ratio of the de Broglie wavelength of electron to the minimum wavelength of X-ray is directly proportional to:
A.$V$
B.$\sqrt V$
C.$V^{\dfrac{3}{2}}$
D.$ V^{\dfrac{7}{2}}$

Answer 1

Hint: The question involves the ratio of the two forms of de Broglie equations, namely the electron or particle and x-ray or the photon. For x-ray$\lambda_{x}=\dfrac{hc}{eV}$ and for electrons$\lambda_{e}=\dfrac{h}{\sqrt{2meV}}$. Also that the kinetic energy $E=eV$ is in both the cases.

Formula used: $\lambda_{x}=\dfrac{hc}{eV}$
$\lambda_{e}=\dfrac{h}{{\sqrt {2meV}}}$

Complete step-by-step solution -
The ratio of the de Broglie wavelength of electron to the minimum wavelength of X-ray is given by the relationship between wavelength $\lambda$ and kinetic energy $E$ is different for particles and photons.
For x-ray:$E=hf=\dfrac{hc}{\lambda_{x}}$ where $E=eV$ ie the kinetic energy of the electrons $e$ accelerated through a potential difference (voltage) $V$, $h$ is the Planck’s constant and $c$ is the speed of light.
Thus can also be written as: $eV= \dfrac{hc}{\lambda_{x}}$ or $\lambda_{x}=\dfrac{hc}{eV}$
For electron: $E=eV=\dfrac{1}{2}\times mv^{2}=\dfrac{p^{2}}{2m}$ or $p=\sqrt{2mE}$ where $p$ is momentum of the electron and $m$ is the mass of the electron. And $\lambda_{e}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}= \dfrac{h}{\sqrt{2meV}}$
Thus taking the ratio
$\dfrac{\lambda_{e}}{\lambda_{x}}=\dfrac{\dfrac{h}{\sqrt{2meV}}}{\dfrac{hc}{eV}}$
$\dfrac{\lambda_{e}}{\lambda_{x}}= \dfrac{h}{\sqrt{2meV}}\times \dfrac{eV}{hc}$
$\dfrac{\lambda_{e}}{\lambda_{x}}\propto\dfrac{V}{\sqrt V}$
$\dfrac{\lambda_{e}}{\lambda_{x}}\propto \sqrt V$
Hence, The ratio of the de Broglie wavelength of electron to the minimum wavelength of X-ray is directly proportional to B.$\sqrt V$

Additional Information:
In 1924 a French physicist Louis de Broglie, gave the De Broglie wavelength, is given by the wave-particle duality, where the wavelength of all the objects in quantum mechanics, is given by the probability density of finding the object at a given any point of the space. The de Broglie wavelength of a particle is inversely proportional to its momentum. Matter waves were first experimentally confirmed to occur in George Paget Thomson's cathode ray diffraction experiment and the Davisson-Germer experiment for electrons, and the de Broglie hypothesis has been confirmed for other elementary particles.

Note: The de Broglie wavelength is expressed in two forms, one for photons and another of particles. Also be careful when taking the ratio between two terms. Be cautious while simplifying the ratios. Note that the kinetic energy $E=eV$ is in both the cases. . For electrons, de Broglie wavelength $\lambda_{e}\propto \dfrac{1}{\sqrt m}$ and$\lambda_{e}\propto \dfrac{1}{\sqrt V}$.
For X-ray, de Broglie wavelength $\lambda_{x}\propto \dfrac{1}{V}$