Question

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
$$\begin{gathered} A.{\lambda }_B={\displaystyle \frac{{\lambda }_G}{3}} \\ B.{\lambda }_B={\displaystyle \frac{{\lambda }_G}{2}} \\ C.{\lambda }_B=2{\lambda }_G \\ D.{\lambda }_B=3{\lambda }_G \end{gathered}$$

Answer 1

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- $\lambda ={\displaystyle \frac{h}{mv}},mvr={\displaystyle \frac{nh}{2\pi }},r=a_0{\displaystyle \frac{n^2}{Z}}$

Complete Step-by-Step solution:
We know that
De- Broglie wavelength:
$\lambda ={\displaystyle \frac{h}{mv}}$ ----- (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:
$mvr={\displaystyle \frac{nh}{2\pi }}$
Here, n is the state of the orbital of an electron.
This equation can be modified as:
$$\begin{gathered} \because mvr={\displaystyle \frac{nh}{2\pi }} \\ \Rightarrow {\displaystyle \frac{h}{mv}}={\displaystyle \frac{2\pi r}{n}}..........(2) \end{gathered}$$
Substitute the value of equation (2) in above equation (1), we get
$\lambda ={\displaystyle \frac{2\pi r}{n}}..........(3)$
As, we know that the radius of atom is given as:
$r=a_0{\displaystyle \frac{n^2}{Z}}$
Here Z is the atomic number of atoms.
Substituting the value of radius in equation (3) we get:
$$\begin{gathered} \because \lambda ={\displaystyle \frac{2\pi r}{n}} \\ \Rightarrow \lambda ={\displaystyle \frac{2\pi a_{0}r{n}^2}{nZ}} \\ \Rightarrow \lambda ={\displaystyle \frac{2\pi a_{0}rn}{Z}} \end{gathered}$$
As, we have to find the result for the hydrogen atom so, Z=1 as the atomic number of hydrogen is 1.
$\Rightarrow \lambda ={\displaystyle \frac{2\pi a_{0}rn}{1}}$
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
$$\begin{gathered} \Rightarrow {\lambda }_B={\displaystyle \frac{2\pi a_{0}r\left(3\right)}{1}} \\ \Rightarrow {\lambda }_B=6\pi a_{0}r.........(4) \end{gathered}$$
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
$$\begin{gathered} \Rightarrow {\lambda }_G={\displaystyle \frac{2\pi a_{0}r\left(1\right)}{1}} \\ \Rightarrow {\lambda }_G=2\pi a_{0}r..........(5) \end{gathered}$$
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)
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\lambda }_B}{{\lambda }_G}}={\displaystyle \frac{6\pi a_{0}r}{2\pi a_{0}r}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_B}{{\lambda }_G}}=3 \\ \Rightarrow {\lambda }_B=3{\lambda }_G \end{gathered}$$
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.