Question

If the velocity of an electron in Bohr’s first orbit is $2.19 \times {10^6}m{s^{ - 1}}$. What will be the de Broglie wavelength associated with it?
A. $2.19 \times {10^{ - 6}}m$
B. $4.38 \times {10^{ - 6}}m$
C. $3.32 \times {10^{ - 10}}m$
D. $3.32 \times {10^{10}}m$

Answer 1

Hint: The de Broglie wavelength is determined by the flow of electrons through the specific orbit and the velocity of electron flow. The velocity of the electron determines the wavelength at which the electron flows or the wavelength of the moving electron which is present in the first orbit.

Complete step-by-step solution: The velocity of the electron in Bohr’s first orbit is $2.19 \times {10^6}m{s^{ - 1}}$. Therefore, the velocity of the electron or $\left( v \right)$ of the specific electron in the first orbit is given. The de Broglie wavelength of the electron can be calculated using the formula:
$\lambda = \dfrac{h}{{mv}}$
In the given formula the $\lambda $ defines the Broglie wavelength of the given electron, $h$ denotes the Planck’s constant and the $m$ defines the mass of the electron. The de Broglie wavelength and the mass of an electron are the two constant values which can be given with the velocity of the electron $\left( v \right)$ which is already provided. The constant value of the Planck’s constant $\left( h \right)$ is $6.626 \times {10^{ - 34}}J/Hz$ and the constant value of the mass of an electron is $9.1 \times {10^{ - 31}}kg$. As the velocity of the electron is already provided as $\left( v \right)$ with the value $2.19 \times {10^6}m{s^{ - 1}}$. The de Broglie wavelength calculated from these values will be:
$\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\left( {9.1 \times {{10}^{ - 31}}} \right) \times \left( {2.19 \times {{10}^6}} \right)}}$
$ \Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{19.929 \times {{10}^{ - 25}}}}$
$ \Rightarrow \lambda = 0.3325 \times {10^{ - 9}}$
$ \Rightarrow \lambda = 3.325 \times {10^{ - 10}}m$
The calculated de Broglie wavelength for the electron that is involved here is $3.32 \times {10^{ - 10}}m$. This means that the movement of the electron in the first orbital takes place in this wavelength considering the wave nature of the electron.

Hence the correct option is (C).

Note:The electrons have both particle nature and wave nature according to the character of duality. The wave nature of the electron shows that the movement of an electron occurs at the given velocity and the specific wavelength.