The wavelength of the electron is [tex]\fbox{\begin\\1.99 \times {10^{ - 10}}\,{\text{m}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\{\text{1}}{\text{.99}}\,\mathop {\text{A}}\limits^{\text{o}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\2 \times {10^{ - 10}}\,{\text{m}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\{\text{2}}\,\mathop {\text{A}}\limits^{\text{o}}\end{minispace}}[/tex].
Further Explanation:
The de-broglie wavelength of an electron is the ratio of the plank’s constant and the momentum of the electron, which is acting along the direction of motion of electron.
Given:
The mass of the electron is [tex]9.11 \times {10^{ - 28}}\,{\text{g}}[/tex].
The velocity with which the electron is moving is [tex]3.66 \times {10^6}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
The value of the planck's constant is [tex]6.63\times10^{-34}\text{ J}/s[/tex].
Concept:
The quantum mechanics or the quantum theory states that any entity or object can exhibit particle like properties and wave like properties as well. This property of entity or object is known as wave-particle duality.
For example, electron is negatively charged particle with small mass also exhibit wavelike and particle like properties. It means that the electrons undergo diffraction and interference just like light wave and it also undergo Compton scattering in which it acts like a point particle.
The wave particle duality of any entity arises due to the interaction process between the electromagnetic radiation and the atoms.
The wave particle duality states that any object which has mass, either it is a particle or a car, will have de-broglie wavelength and the wavelength is inversely proportional to the momentum of that object.
The de-broglie wavelength of electron is:
[tex]\fbox{\begin\\\lambda=\dfrac{h}{p}\end{minispace}}[/tex] …… (1)
Here, [tex]\lambda[/tex] is the wavelength of the electron, [tex]h[/tex] is the plank’s constant and [tex]p[/tex] is the momentum of the electron.
The momentum of the electron is:
[tex]\fbox{\begin\\p = mv\end{minispace}}[/tex] …… (2)
Here, [tex]m[/tex] is the mass of electron and [tex]v[/tex] is the velocity with which electron is moving.
Convert the mass of electron from grams into kilogram.
[tex]\begin{aligned}\\m&=9.11\times{10^{-28}}\,{\text{g}}\\&=9.11\times{10^{-31}}\,{\text{kg}}\\\end{aligned}[/tex]
Substitute the value of mass and velocity in equation (2).
[tex]\begin{aligned}p&= \left( {9.11 \times {{10}^{ - 31}}\,{\text{kg}}} \right)\left( {3.66 \times {{10}^6}\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right \kern-\nulldelimiterspace} {\text{s}}}} \right)\\&= 3.33 \times {10^{ - 24}}\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}[/tex]
Substitute the value of momentum and plank’s constant in equation (1).
[tex]\begin{aligned}\lambda&=\dfrac{6.63\times10^{-34}\text{ J}/s}{3.33\times10^{-24}\text{ kg}\cdot\text{m}/s}\\&=1.99\times10^{-10}\text{m}\end{aligned}[/tex]
Thus, the wavelength of the electron is [tex]\fbox{\begin\\1.99 \times {10^{ - 10}}\,{\text{m}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\{\text{1}}{\text{.99}}\,\mathop {\text{A}}\limits^{\text{o}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\2 \times {10^{ - 10}}\,{\text{m}}\end{minispace}}[/tex] or [tex]\fbox{\begin\\{\text{2}}\,\mathop {\text{A}}\limits^{\text{o}}\end{minispace}}[/tex].
Learn more:
1.Threshold frequency of cesium https://brainly.com/question/6953278
2. Average translational energy https://brainly.com/question/9078768
3. Conservation of energy brainly.com/question/3943029
Answer Details:
Grade: High School
Subject: Physics
Chapter: Dual nature
Keywords:
De-broglie wavelength, electron, wave particle duality, momentum, planck’s constant, quantum mechanics, quantum theory, 1.99*10-10 m, 1.99*10^-10 m, 1.99 angstorm, 2*10-10 m, 2*10^-10 m, 2 angstorm.
