What is the de Broglie wavelength associated with the hydrogen electron in its second orbit?

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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.