Question

What do you mean by de Broglie principle? Write de Broglie equation. What potential must be applied to an electron microscope to produce an electron beam of wavelength $0.41\times {10}^{-10}m$?

Answer 1

Hint: De Broglie principle talks about duality behaviour of matter. The de Broglie equation relates the momentum of a particle to its wavelength that corresponds to the matter behaving as a wave.

Formula used: $\lambda ={\displaystyle \frac{h}{p}}$, where $\lambda$ is the wavelength of the particle when as it acts as a wave, $h$ is the Planck’s constant and $p$ is the momentum of the particle.

Complete step by step solution:
The de Broglie principle tells us that matter can act as waves just like light can act as waves and particles (photons). So every particle will have a wavelength corresponding to its wave behavior.
The de Broglie equation relates the momentum of a particle with the wavelength of its wave behavior and is written as:
$\Rightarrow \lambda ={\displaystyle \frac{h}{p}}$
Now we want to find the potential that must be applied to produce an electron beam of wavelength $0.41\times {10}^{-10}m$. We should start by finding the kinetic energy (K.E.) gained by an electron when it is accelerated in a potential which is given as the product of the charge of the electron and the potential. Hence the kinetic energy is:
K.E. is equal to $eV$ where $e=1.6\times {10}^{-19}C$ is the charge of the electron and $V$ is the potential applied.
We know that the momentum of a particle is related to its kinetic energy as:
$\Rightarrow p=\sqrt{2m\times \text{K}\text{.E}\text{.}}$
$\Rightarrow \sqrt{2meV}$
where $m$ is the mass of the electron.
Substituting the value of momentum in de Broglie’s equation, we get:
$\Rightarrow \lambda ={\displaystyle \frac{h}{p}}$
$\Rightarrow {\displaystyle \frac{h}{\sqrt{2meV}}}$
Rearranging the equation, we get
$\Rightarrow \sqrt{2meV}={\displaystyle \frac{h}{\lambda }}$
On squaring both sides and then dividing both sides by $2me$ ,
$\Rightarrow V={\displaystyle \frac{h^2}{2me{\lambda }^2}}$
Substituting $\lambda =0.41\times {10}^{-10}m=0\text{.41}\times \text{1}{\text{0}}^{-10}m$, $h=6.636\times {10}^{-34}$ and $m=9.1\times {10}^{-31}$, we get:
$V={\displaystyle \frac{{(6.636\times {10}^{-34})}^2}{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times {(0.41\times {10}^{-10})}^2}}$
Solving further we get,
$\therefore V=900\text{Volts}$
Hence, we will require 900 Volts of potential to accelerate an electron so that it will have a wavelength of $0.41\times {10}^{-10}m$.

Additional Information:
The de Broglie principle has been proved using experiments like double-slit interference of an electron beam which is only possible if the electrons, which are particles, also have a wave-like behavior since interference patterns can only be formed by the superposition of waves.

Note:
We must be careful to not balance the energy of the electron beam with the energy gain due to the potential since the energy stored by the electron is partly stored in the form of kinetic energy. So,
${\displaystyle \frac{hc}{\lambda }}\ne eV$