Question

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5{{a}_{0}}$ where ${{a}_{0}}$ the Bohr radius is. Its orbital angular momentum is $\dfrac{3h}{2\pi }$ . It is given that $h$ is Planck's constant and $R$ is Rydberg constant. The possible wavelength(s), if the atom excited will be,
There may be multiple correct answers for this question.
$\begin{aligned} & A.\dfrac{9}{32R} \\\ & B.\dfrac{9}{16R} \\\ & C.\dfrac{9}{5R} \\\ & D.\dfrac{4}{3R} \\\ \end{aligned}$

Answer 1

First of all find the $n$ and $Z$ value using the equation for the radius of the orbit the possible wavelength can be determined by the equation,
$r={{a}_{0}}\dfrac{{{n}^{2}}}{Z}$
Then find out the possible wavelength using the equation,
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left[ \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right]$
Where $R$ be the Rydberg constant, ${{n}_{1}}$ and ${{n}_{2}}$ be the level of orbital or the position of orbital and $Z$ be the atomic number. Chances are there for multiple correct answers.

Complete step-by-step answer:
It is already given that, the radius of orbit of an electron in a Bohr orbit is,
$r=4.5{{a}_{0}}$
The radius of the Bohr orbit is generally given as,
$r={{a}_{0}}\dfrac{{{n}^{2}}}{Z}$
Therefore we can equate both these equations together,
${{a}_{0}}\dfrac{{{n}^{2}}}{Z}=4.5{{a}_{0}}$
And also we can write that,
$\dfrac{nh}{2\pi }=\dfrac{3h}{2\pi }$
So from this we can write that,
$n=3$ and $Z=2$
Therefore now we can find the possible wavelength which is given by the equation,
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left[ \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right]$
Here three types of transition are possible, one can be from first state to third state.
This can be found using the above mentioned equation.
Substituting the values in it,

& \dfrac{1}{{{\lambda }_{1}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\\ & {{\lambda }_{1}}=\dfrac{9}{32R} \\\ \end{aligned}$$ Another possibility is that the transition can take place from second state to the first one, $$\begin{aligned} & \dfrac{1}{{{\lambda }_{2}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right] \\\ & {{\lambda }_{2}}=\dfrac{1}{3R} \\\ \end{aligned}$$ And the last possibility is that, the transition may happens from second state to the third state, $$\begin{aligned} & \dfrac{1}{{{\lambda }_{3}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\\ & {{\lambda }_{3}}=\dfrac{9}{5R} \\\ \end{aligned}$$ In the given options, there are two options which are having these found out values in it. **So, the correct answers are “Option A and C”.** **Note:** In physics, the Rydberg constant is defined as a physical constant which is in relation with atomic spectra. It is abbreviated as $ R $. Rydberg constant was firstly found from the Rydberg formula as a mere constant. Later Niels Bohr developed it from the fundamental constants.