Question

The radius of the orbit of an electron in a Hydrogen-like atom is 4.5${{a}_{\circ }}$, where ${{a}_{\circ }}$ is the Bohr radius. Its orbital angular momentum is $\dfrac{3h}{2\pi }$. It is given that h is Planck’s constant and R is the Rydberg constant. The possible wavelengths, when the atom de-excited is (are)
(A)$\dfrac{9}{32R}$
(B)$\dfrac{9}{16R}$
(C)$\dfrac{9}{5R}$
(D)$\dfrac{4}{3R}$

Answer 1

We use three formulas to solve this problem. Firstly we use the formula of radius of Hydrogen-like species given by Neils Bohr and the formula for orbital angular momentum. Then we use the Balmer’s formula involving wavelengths of Hydrogen-like species. For questions as these were there maybe be more than one possible wavelength, we need to check the options carefully.
Formula Used:
The radius of Hydrogen and Hydrogen-like species is given by
r = ${{a}_{0}}\dfrac{{{n}^{2}}}{Z}$
The formula for orbital angular momentum is given by
L = $\dfrac{nh}{2\pi }$
The formula given by Balmer for hydrogen-like species is given as
$\dfrac{1}{\lambda }=R{{Z}^{2}}(\dfrac{1}{{{n}^{2}}}-\dfrac{1}{{{m}^{2}}})$

Complete answer:
Given,
Radius of Hydrogen-like species = 4.5${{a}_{\circ }}$
Orbital angular momentum = $\dfrac{3h}{2\pi }$
To find: The possible wavelengths when the atom de-excites
We have,
Using the radius formula given by Bohr
4.5${{a}_{0}}$ = ${{a}_{0}}$ $\dfrac{{{n}^{2}}}{Z}$ ............. (1)
The orbital angular momentum is given by
$\dfrac{3h}{2\pi }$ = $\dfrac{nh}{2\pi }$ ............... (2)
Where n is the number of state
Z is the number of protons or the atomic number
Solving (1) and (2), we get
n = 3
Z = 2
That is, it is a hydrogen-like species in the 3rd excited state with atomic number 2.
Now, we plug in these values of n and Z in the Balmer formula to get the possible wavelengths
In the Balmer formula, m > n
So, we get, for n = 1 and m = 2
$\begin{aligned} & \dfrac{1}{{{\lambda }_{1}}}=R{{(2)}^{2}}(\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}}) \\\ & \Rightarrow {{\lambda }_{1}}=\dfrac{1}{3R} \\\ \end{aligned}$
Again, for n= 1, m = 3
$\begin{aligned} & \dfrac{1}{{{\lambda }_{2}}}=R{{(2)}^{2}}(\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}}) \\\ & \Rightarrow {{\lambda }_{2}}=\dfrac{9}{32R} \\\ \end{aligned}$
Again, for n = 2, m = 3
$\begin{aligned} & \dfrac{1}{{{\lambda }_{3}}}=R{{(2)}^{2}}(\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}}) \\\ & \Rightarrow {{\lambda }_{3}}=\dfrac{9}{5R} \\\ \end{aligned}$

Checking for the possible wavelengths in this manner, we get the two possible wavelengths as (A) $\dfrac{9}{32R}$ and (C) $\dfrac{9}{5R}$ .

Additional Information:
The Balmer formula for calculation of wavelength can be used to calculate different types of spectral lines identified as Lyman, Balmer, Paschen, Brackett and Pfund which are classified on the basis of their n number. For every n, m can take all values greater than n to get the series of spectral lines.

Note:
In this given problem, the steps must be followed in the given order. We need to calculate the n and Z number first and then move into finding the possible wavelengths of the Hydrogen-like species(Helium here) with the help of the Balmer formula. The formulas of radius and orbital angular momentum must necessarily be kept in mind.