Question

The de-Broglie wavelength of an electron in ${4^{{\text{th}}}}$ orbit is _____. ($r$ is radius of ${1^{{\text{st}}}}$ orbit)
(A) $\dfrac{{\pi r}}{2}$
(B) $4\pi r$
(C) $8\pi r$
(D) $16\pi r$

Answer 1

First of all, we will use the equation for de-Broglie wavelength and then we will modify the equation of the angular momentum of an electron which corresponds to the order of orbit. We will manipulate the equation accordingly and obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
An electron is revolving around the nucleus in a particular orbit.
The radius of the first orbit is given as $r$ .
We are asked to find the wavelength of the electron in the fourth orbit.
To begin with, let us know something about a moving electron. An electron which is moving in a particular orbit has both the particle nature and wave nature associated with it. As electrons have a rest mass, they have a de Broglie wavelength that is very small, about $0.01$ nanometres for speeds that can be quickly reached, unlike photons. This means that many smaller objects can be observed through a microscope using electron "matter waves" rather than photon light waves.
To proceed further, we now use the expression which gives the de-Broglie wavelength, which is given below:
$\lambda = \dfrac{h}{{mv}}$ …… (1)
Where,
$\lambda$ indicates the wavelength.
$h$ indicates Planck’s constant.
$m$ indicates mass of the electron.
$v$ indicates velocity of the electron.
According to Bohr’s theory, we have the angular momentum of an electron, which is given by:
$mvr = \dfrac{{nh}}{{2\pi }}$ …… (2)
Where,
$v$ indicates the velocity of the electron in that particular orbit.
$r$ indicates the radius of that particular orbit.
$n$ indicates a particular orbit.
From equation (2), we can write as:
$mv = \dfrac{{nh}}{{2\pi r}}$ …… (3)
Now, we use the equation (3) in equation (1) and we get:
$\lambda = \dfrac{h}{{mv}} \\\ \Rightarrow \lambda = \dfrac{h}{{\left( {\dfrac{{nh}}{{2\pi r}}} \right)}} \\\ \Rightarrow \lambda = \dfrac{{2\pi rh}}{{nh}} \\\ \Rightarrow \lambda = \dfrac{{2\pi r}}{n} \\\$
Since, we are asked to find the wavelength of the electron in the ${4^{{\text{th}}}}$ orbit. So, we can write:
$\lambda = \dfrac{{2\pi r}}{4} \\\ \therefore \lambda = \dfrac{{\pi r}}{2} \\\$
Hence, the wavelength of the electron in the fourth orbit is $\dfrac{{\pi r}}{2}$ .

The correct option is A.

Note: This problem is based on the model of an atom. While solving this problem, always remember that angular momentum of an electron always increases with the increase in the order of the orbit. Hence, the wavelength decreases as the electron jumps to a higher orbit.