The energy corresponding to one of the lines in the Paschen series for $H$atoms is $18.16 \times {10^{ - 20}}{\text{J}}$. Find the quantum numbers for transition which produce this line.
Answer
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Hint: To answer this question, you should recall the concept of spectral lines of a hydrogen atom. Bohr’s theory and the various transition series when an excited electron comes back to the lower energy state. Once the electrons in the gas are excited, they make transitions between the energy levels.
The formula used:
\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{c}}{{\text{m}}^{{\text{ - 1}}}}\]where \[R\]is Rydberg constant, \[n\]is the shell and \[\dfrac{1}{\lambda }\]is the wavenumber, \[R = 109678{\text{c}}{{\text{m}}^{{\text{ - 1}}}}\]
Complete step by step answer:
For the Paschen series of the hydrogen atom, \[{n_1} = 3\].
We know that energy is given by: \[E = \dfrac{{hc}}{\lambda }\].
From here we can find the value of wavenumber
\[E = 18.16 \times {10^{ - 20}}{\text{J}}\]
\[ \Rightarrow \dfrac{1}{\lambda } = 913571.794{{\text{m}}^{{\text{ - 1}}}}\]
Substituting the value:
\[\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}}{\text{ }} - \dfrac{1}{{{n_2}^{2}}}} \right)\;\]
Substituting the values, we have:
\[ \Rightarrow 913572.794 = 109678 \times {(1)^2}\left( {\dfrac{1}{{{3^2}}} - {\text{ }}\dfrac{{1}}{{{n^2}}}} \right)\]
Solving and rearranging:
\[ \Rightarrow \dfrac{1}{{100}} \times 8.3296 = \dfrac{1}{9} - \dfrac{1}{{{n^2}}}\].
\[ \Rightarrow {n^2} = 35.95155\].
Ultimately, we will have:
\[ \Rightarrow n \cong 6\].
Hence, we can conclude that the principal quantum number for this electron is = 6
Note:
You should know that four quantum numbers can be used to completely describe all the attributes of a given electron belonging to an atom, these are: Principal quantum number, denoted by\[n\]; Orbital angular momentum quantum number (or azimuthal quantum number), denoted by\[\;l\]; Magnetic quantum number, denoted by \[{m_l}\]; The electron spin quantum number, denoted by \[{m_s}\]. The spectral series are classified as:
– Lyman series
– Balmer series
– Paschen series
– Bracket series
– Pfund series
The formula used:
\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right){\text{c}}{{\text{m}}^{{\text{ - 1}}}}\]where \[R\]is Rydberg constant, \[n\]is the shell and \[\dfrac{1}{\lambda }\]is the wavenumber, \[R = 109678{\text{c}}{{\text{m}}^{{\text{ - 1}}}}\]
Complete step by step answer:
For the Paschen series of the hydrogen atom, \[{n_1} = 3\].
We know that energy is given by: \[E = \dfrac{{hc}}{\lambda }\].
From here we can find the value of wavenumber
\[E = 18.16 \times {10^{ - 20}}{\text{J}}\]
\[ \Rightarrow \dfrac{1}{\lambda } = 913571.794{{\text{m}}^{{\text{ - 1}}}}\]
Substituting the value:
\[\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_1}^2}}{\text{ }} - \dfrac{1}{{{n_2}^{2}}}} \right)\;\]
Substituting the values, we have:
\[ \Rightarrow 913572.794 = 109678 \times {(1)^2}\left( {\dfrac{1}{{{3^2}}} - {\text{ }}\dfrac{{1}}{{{n^2}}}} \right)\]
Solving and rearranging:
\[ \Rightarrow \dfrac{1}{{100}} \times 8.3296 = \dfrac{1}{9} - \dfrac{1}{{{n^2}}}\].
\[ \Rightarrow {n^2} = 35.95155\].
Ultimately, we will have:
\[ \Rightarrow n \cong 6\].
Hence, we can conclude that the principal quantum number for this electron is = 6
Note:
You should know that four quantum numbers can be used to completely describe all the attributes of a given electron belonging to an atom, these are: Principal quantum number, denoted by\[n\]; Orbital angular momentum quantum number (or azimuthal quantum number), denoted by\[\;l\]; Magnetic quantum number, denoted by \[{m_l}\]; The electron spin quantum number, denoted by \[{m_s}\]. The spectral series are classified as:
– Lyman series
– Balmer series
– Paschen series
– Bracket series
– Pfund series
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