
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{align}
& A.\dfrac{9}{32R} \\
& B.\dfrac{9}{16R} \\
& C.\dfrac{9}{5R} \\
& D.\dfrac{4}{3R} \\
\end{align} $
Answer
570.9k+ views
Hint: 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,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{1}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& {{\lambda }_{1}}=\dfrac{9}{32R} \\
\end{align}\]
Another possibility is that the transition can take place from second state to the first one,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{2}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right] \\
& {{\lambda }_{2}}=\dfrac{1}{3R} \\
\end{align}\]
And the last possibility is that, the transition may happens from second state to the third state,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{3}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& {{\lambda }_{3}}=\dfrac{9}{5R} \\
\end{align}\]
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.
$ 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,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{1}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& {{\lambda }_{1}}=\dfrac{9}{32R} \\
\end{align}\]
Another possibility is that the transition can take place from second state to the first one,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{2}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right] \\
& {{\lambda }_{2}}=\dfrac{1}{3R} \\
\end{align}\]
And the last possibility is that, the transition may happens from second state to the third state,
\[\begin{align}
& \dfrac{1}{{{\lambda }_{3}}}=R{{Z}^{2}}\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& {{\lambda }_{3}}=\dfrac{9}{5R} \\
\end{align}\]
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.
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