Question

If the velocity of the electron in Bohr’s first orbit is $2.19\times 1{{0}^{6}}m{{s}^{-1}}$ , calculate the de Broglie wavelength associated with it.

Answer 1

Hint: While studying quantum mechanics, the de Broglie wavelength is an important concept. The wavelength which is associated with an entity in relation to its mass and momentum is known as the de Broglie wavelength. A particle’s de Broglie wavelength is normally inversely proportional to its force.

Complete step by step answer:
- As we know, matter has a dual nature of wave-particles. The de Broglie waves was named after Louis de Broglie .It is the property of an object that varies in space or time while behaving similar to waves and it’s also called matter-waves.
- In the case of electrons which go in circles around the nucleus in atoms, the de Broglie waves normally exist as a closed-loop, so that they can exist only as standing waves. For this condition, the electrons in atoms will circle nuclei in particular states, or configurations and they are usually called as stationary orbits.

- The De-Broglie wavelength can be calculated by using the following formula:
\[\lambda =\dfrac{h}{mv}\]
Where $\lambda $ is the wavelength of electron
             h is the Planck’s constant ( $6.626\times {{10}^{-34}}Js$)
             m is the mass of the electron( $9.1\times {{10}^{-31}}kg$)
             $v$ is the velocity of the electron (given as $2.19\times 1{{0}^{6}}m{{s}^{-1}}$)
Substitute this values in the above equation, we get
\[\lambda =\dfrac{6.626\times {{10}^{-34}}Js}{9.1\times {{10}^{-31}}kg\times 2.19\times 1{{0}^{6}}m{{s}^{-1}}}=0.03311\times {{10}^{-9}}m=3.311\overset{\text{o}}{\mathop{\text{A}}}\,\]

  Therefore the de Broglie wavelength associated with the electron in Bohr’s first orbit is $3.311\overset{\text{o}}{\mathop{\text{A}}}\,$or $3.3\times {{10}^{-10}}m$.

Note: Keep in mind that the de Broglie wavelength is inversely proportional to momentum since momentum is obtained by multiplying mass by velocity. In order to find the de Broglie wavelength in terms of kinetic energy we could use the equation $\lambda =\dfrac{h}{\sqrt{2m{{E}_{k}}}}$ where ${{E}_{k}}$ is the kinetic energy of electron.