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**Hint:**The de Broglie wavelength is calculated by dividing the Planck's constant with the momentum of the electron. The radius of the nth orbital is equal to the square of n multiplied by the radius of the first orbit.

**Complete answer:**Let us understand the de Broglie wavelength:

The wavelength of the wave associated with any material particle was calculated by analogy with photons as follows:

In the case of a photon, if it is assumed to have wave character, its energy is given by,

\[E=h\text{v}\]

Where v is the frequency of the wave and h is the planck's constant.

If the photon is supposed to have a particle nature, its energy is,

\[\text{E=}m{{c}^{2}}\]

Where m is the mass of the particle and c is the speed of light.

On solving both the equation, we get

\[h\text{v}=m{{c}^{2}}\]

\[but\text{ v=}\dfrac{c}{\lambda }\]

\[h\dfrac{c}{\lambda }=m{{c}^{2}}\text{ or }\lambda =\dfrac{h}{mc}\]

For applying it on any material, the c is replaced with v (velocity of the material).

$\lambda =\dfrac{h}{m\text{v}}$ .......(i)

This is called the de Broglie equation and \[\lambda \]is the de Broglie wavelength.

The radii of the ground states of the hydrogen atom can be expressed by,

\[{{r}_{n}}={{n}^{2}}{{r}_{0}}\]

Where\[{{r}_{0}}\] is the radius of the first stationary state and is called the Bohr radius.

Now, according to question the radius of the 3rd orbit is equal to

\[{{r}_{3}}={{3}^{2}}x\]

Given in the question is ‘x’. Hence the radius of 3rd orbit is

\[{{r}_{3}}=9x\]

Now, according to Bohr’s postulate of angular momentum,

\[mvr=n\dfrac{h}{2\pi }\]

Putting the values in the above equation, we get

\[mv9x=3\dfrac{h}{2\pi }\]

On solving the above equation,

\[\dfrac{2\pi \text{ x }9x}{3}=\dfrac{h}{m\text{v}}\]

\[\dfrac{h}{m\text{v}}=6\pi x\]

$\lambda =6\pi x$ .........[From equation (i)]

Hence the de Broglie wavelength is \[6\pi x\]

**So, the correct answer is “Option B”.**

**Note:**All the formulas and equations should be used correctly. The significance of the de Broglie equation lies in the fact that it relates to the particle character with the wave character of matter. The de Broglie equation applies to all objects.

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