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In the H-atom if x is the radius of the first Bohr orbit, the De Broglie wavelength of an electron in $3rd$ orbit is:
A. $3\pi x$
B. $6\pi x$
C. $\dfrac{{9x}}{2}$
D. $\dfrac{x}{2}$

Last updated date: 24th Jul 2024
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Hint: In the question, we have to calculate the De-Broglie wavelength. Using the Bohr postulates, the relation was determined for the radius of Bohr orbit.
We have, $mvr = \dfrac{{nh}}{{2\pi }}$ ; v represents the velocity of an electron, r represents the radius, h represents Planck’s constant, n represents the principal quantum number of orbit.
Here, the mentioned expression is being used for angular momentum given by De-Broglie.
Further the general formula was determined for radius that can be written as $r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZK{e^2}}}$

Complete step by step answer:
As we already mentioned the general formula the radius of ${n^{th}}$ Bohr orbit i.e.
$r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZK{e^2}}}$
n represents the principal quantum number of orbits like n is 0, 1, 2, …
h represents the Planck’s constant
m represents the mass of an electron
K represents the Coulomb’s constant
e represents the charge of an electron
In this expression Z represents the atomic number.
Now, we know that the atomic number of hydrogen is 1.
Therefore, the expression of radius for hydrogen atom will be as follows:
$r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mK{e^2}}}$
So, we can say that according to the Bohr $r\alpha {n^2}$
We are given that x is the radius of Bohr’s first orbit (${r_1}$), thus using the above expression we can write it as follows:
$\dfrac{{{r_1}}}{{{r_3}}} = \dfrac{{{n_1}^2}}{{{n_3}^2}}$ Here ${r_3}$ represents the radius of $3rd$ orbit.
$ \Rightarrow \dfrac{{{r_1}}}{{{r_3}}} = \dfrac{{{1^2}}}{{{3^2}}}$
$ \Rightarrow {r_3} = 9{r_1}$ (Solving it in terms of ${r_1}$)
Here we know ${r_1}$ is x.
So, ${r_3} = 9x$
Now, if we see the expression given by De-Broglie for angular momentum i.e.
$mvr = \dfrac{{nh}}{{2\pi }}$
If the electron is in $3rd$ orbit then $mv{r_3} = \dfrac{{3h}}{{2\pi }}$
$ \Rightarrow \dfrac{h}{{mv}} = \dfrac{{2\pi {r_3}}}{3}$
We know that De-Broglie have given the expression for wavelength which can be written as:
$\lambda = \dfrac{h}{{mv}}$ ; $\lambda $ is the De-Broglie wavelength
Thus, $\lambda = \dfrac{{2\pi {r_3}}}{3}$
After substituting the value of ${r_3}$ in the above expression, we get
$\lambda = \dfrac{{2\pi 9x}}{3}$
$ \Rightarrow \lambda = 6\pi x$
In the last, we can conclude that the wavelength for electron in $3rd$ orbit is $6\pi x$

So, the correct answer is Option B.

Note: We have come across the term ‘angular momentum of electron’. The concept of angular momentum is given by De-Broglie. The electron revolving around the nucleus shows both wave and particle character. The circumference of the orbit is considered to be an integral multiple of the wavelength of the electron whose wave is existing in the phase.