The de-Broglie wavelength $\left( {{\lambda _B}} \right)$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $\left( {{\lambda _G}} \right)$ by
\[
  A.{\text{ }}{\lambda _B} = \dfrac{{{\lambda _G}}}{3} \\
  B.{\text{ }}{\lambda _B} = \dfrac{{{\lambda _G}}}{2} \\
  C.{\text{ }}{\lambda _B} = 2{\lambda _G} \\
  D.{\text{ }}{\lambda _B} = 3{\lambda _G} \\
 \]

Hint: In order to deal with this question first we will write de- Broglie wavelength equation then we will proceed further by putting this value in conservation of angular momentum formula. After using these two formulas, we will get a relation in terms of radius of atom. Further we will use the formula for the radius of an atom in terms of atomic number to find the relation.
Formula used- $\lambda = \dfrac{h}{{mv}},mvr = \dfrac{{nh}}{{2\pi }},r = {a_0}\dfrac{{{n^2}}}{Z}$

Complete Step-by-Step solution:
We know that
De- Broglie wavelength:
$\lambda = \dfrac{h}{{mv}}$ ----- (1)
Where h is the planck's constant, m is the mass and v is the velocity.
Also, we know that conservation of angular momentum is given as:
$mvr = \dfrac{{nh}}{{2\pi }}$
Here, n is the state of the orbital of an electron.
This equation can be modified as:
$
  \because mvr = \dfrac{{nh}}{{2\pi }} \\
   \Rightarrow \dfrac{h}{{mv}} = \dfrac{{2\pi r}}{n}..........(2) \\
 $
Substitute the value of equation (2) in above equation (1), we get
$\lambda = \dfrac{{2\pi r}}{n}..........(3)$
As, we know that the radius of atom is given as:
$r = {a_0}\dfrac{{{n^2}}}{Z}$
Here Z is the atomic number of atoms.
Substituting the value of radius in equation (3) we get:
$
  \because \lambda = \dfrac{{2\pi r}}{n} \\
   \Rightarrow \lambda = \dfrac{{2\pi {a_0}r{n^2}}}{{nZ}} \\
   \Rightarrow \lambda = \dfrac{{2\pi {a_0}rn}}{Z} \\
 $
As, we have to find the result for the hydrogen atom so, Z=1 as the atomic number of hydrogen is 1.
$ \Rightarrow \lambda = \dfrac{{2\pi {a_0}rn}}{1}$
As, $\left( {{\lambda _B}} \right)$ is the De-Broglie wavelength correlated with the electron orbiting the hydrogen atom in the second excited state.
So, for $\left( {{\lambda _B}} \right)$ , the value n=3
$
   \Rightarrow {\lambda _B} = \dfrac{{2\pi {a_0}r\left( 3 \right)}}{1} \\
   \Rightarrow {\lambda _B} = 6\pi {a_0}r.........(4) \\
 $
Also we know $\left( {{\lambda _G}} \right)$ is the De-Broglie wavelength associated with the electron orbiting in the ground state of a hydrogen atom.
So, for $\left( {{\lambda _G}} \right)$ , the value n=1
$
   \Rightarrow {\lambda _G} = \dfrac{{2\pi {a_0}r\left( 1 \right)}}{1} \\
   \Rightarrow {\lambda _G} = 2\pi {a_0}r..........(5) \\
 $
Now, let us compare $\left( {{\lambda _B}} \right)$ and $\left( {{\lambda _G}} \right)$ to find the relation between them.
Let us divide equation (4) by equation (5)
$
   \Rightarrow \dfrac{{{\lambda _B}}}{{{\lambda _G}}} = \dfrac{{6\pi {a_0}r}}{{2\pi {a_0}r}} \\
   \Rightarrow \dfrac{{{\lambda _B}}}{{{\lambda _G}}} = 3 \\
   \Rightarrow {\lambda _B} = 3{\lambda _G} \\
 $
Hence, the relation between the De-Broglie wavelengths associated with the ground state and the second excited state of the hydrogen atom is ${\lambda _B} = 3{\lambda _G}$ .
So, the correct answer is option D.

Note- The De Broglie wavelength is, according to wave-particle duality, a wavelength manifested in all objects in quantum mechanics which determines the probability density of finding the object at a given point in the space of the configuration. The diameter of a de Broglie particle is inversely proportional to its momentum.