Question

Dual behavior of matter proposed by de Broglie led to the discovery of an electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is $1.6 \times 10^6 m/s$. Calculate de Broglie wavelength associated with this electron.

Answer 1

Hint: The concept of dual nature of particles proposed by de Broglie is used in an electron microscope and de Broglie equation is used to calculate the de Broglie wavelength associated with an electron.

Formula: $\lambda = \dfrac{h}{{mv}}$, where mass of electron, $ m = 9.1\mathop { \times 10}\nolimits^{ - 31} kg$; v is the velocity of electron and $h = 6.626 \times \mathop {10}\nolimits^{ - 34} Js$.

Complete step by step solution:
There is a concept in quantum mechanics which states that every particle can be described either as a particle or a wave. Also, it expresses the inability of the classical concepts, “particle” or “wave” to fully describe the behavior of matter.
As we are already aware that dual behavior of matter means that matter at atomic scale in some circumstances behaves as particles while in some, they behave like waves.
This means that some phenomena like photoelectric effect and black body radiation could only be explained by taking into account the particle nature of a matter while others could only be explained by taking into consideration the wave nature of material particles.
After so many theories, de-Broglie suggested the theory of “dual nature of radiation”.
According to de-Broglie theory:
Every object in motion should have wave character.
Wavelength of objects having larger masses cannot be detected whereas the wavelength of particles having very small masses (like electrons and other subatomic particles).
Derivation of de Broglie equation:
We know that frequency, $f = \dfrac{c}{\lambda }$.
Also, $hf = m{c^2}$.
This implies that de Broglie wavelength, $\lambda = \dfrac{h}{{mc}}$.
If the velocity of a particle is v i.e. $c = v$ then, $\lambda = \dfrac{h}{{mv}}$.
Now coming to the question in which we have to find de Broglie wavelength and given velocity of electron, $v = 1.6 \times \mathop {10}\nolimits^6 m{s^{ - 1}}$.
Calculation:
$
  h = 6.626 \times \mathop {10}\nolimits^{ - 34} Js \\
  m = 9.1\mathop { \times 10}\nolimits^{ - 31} kg \\
  v = 1.6 \times \mathop {10}\nolimits^6 m{s^{ - 1}} \\
  \lambda = \dfrac{h}{{mv}} = \dfrac{{6.626 \times \mathop {10}\nolimits^{ - 34} }}{{\left( {9.1 \times \mathop {10}\nolimits^{ - 31} } \right)\left( {1.6 \times \mathop {10}\nolimits^6 } \right)}} = 4.55 \times \mathop {10}\nolimits^{ - 10} m \\
$.
Therefore, de Broglie wavelength of electron having $v = 1.6 \times \mathop {10}\nolimits^6 m{s^{ - 1}}$ is $ 4.55 \times \mathop {10}\nolimits^{ - 10} m.$.

Note: It should be remembered that an electron microscope works by using an electron beam instead of visible light and an electron detector instead of our eyes.