Calculate the shortest and longest wavelengths in the hydrogen spectrum of Lyman series.
Answer
561.2k+ views
Hint: The lines arise when electron comes from higher energy level to lower energy level of Principle quantum number one is known as Lyman series. Longest wavelength corresponds to the lowest energy transition and shortest wavelength corresponds to the highest energy transition.
Formula used:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Complete step by step answer:
The formula to determine the wavelength of transitions in hydrogen spectrum is as follows:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Where,
\[\lambda\] is the wavelength.
\[{{\text{R}}_{\text{H}}}\] is the Rydberg constant.
The value of Rydberg constant is $109678\,{\text{c}}{{\text{m}}^{ - 1}}$ .
\[{{\text{n}}_{{\text{lower}}}}\] is the principle quantum of the energy level in which the electron comes.
\[{{\text{n}}_{{\text{higher}}}}\] is the principle quantum of the energy level from which the electron comes.
Lyman series arise when electrons from higher energy levels come to the energy level of principle quantum number one. So, the \[{{\text{n}}_{{\text{higher}}}}\] is \[{{\text{n}}_1}\].
So, the wavelength formula for Lyman series is as follows:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Shortest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the highest energy level. The highest energy level is \[{{\text{n}}_\infty }\].
So, the shortest wavelengths is,
Substitute $109678\,{\text{c}}{{\text{m}}^{ - 1}}$ for Rydberg constant, $n_1=1$ and \[{{\text{n}}_\infty }\] for \[{{\text{n}}_{{\text{higher}}}}\].
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_\infty ^2}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\]
\[\Rightarrow \lambda = \dfrac{{\text{1}}}{{\,109678\,{\text{c}}{{\text{m}}^{ - 1}}}}\]
\[\Rightarrow \lambda = 9.11 \times \,{10^{ - 6}}\,{\text{cm}}\]
So, the shortest wavelength is \[9.11 \times \,{10^{ - 6}}\,{\text{cm}}\].
Longest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the next higher energy level. The next higher energy level is second so, the \[{{\text{n}}_{{\text{higher}}}}\] is \[2\].
So, the longest wavelengths is,
Substitute \[{\text{c}}{{\text{m}}^{ - 1}}\] for Rydberg constant and \[{{\text{n}}_2}\] for \[{{\text{n}}_{{\text{higher}}}}\].
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^2}} - \dfrac{{\text{1}}}{{{\text{n}}_2^2}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\left[ {\dfrac{{\text{1}}}{{{1^2}}} - \dfrac{{\text{1}}}{{{2^2}}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}} \times \dfrac{3}{4}\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,82258.5\,{\text{c}}{{\text{m}}^{ - 1}}\]
\[\Rightarrow \lambda = \dfrac{{\text{1}}}{{82258.5\,{\text{c}}{{\text{m}}^{ - 1}}}}\]
\[\Rightarrow \lambda = 1.22 \times {10^{ - 5}}\,{\text{cm}}\]
So, the longest wavelength is \[1.22 \times {10^{ - 5}}\,{\text{cm}}\].
Therefore, the shortest and longest wavelengths in hydrogen spectrum of Lyman series are \[9.11 \times \,{10^{ - 6}}\,{\text{cm}}\]and \[1.22 \times {10^{ - 5}}\,{\text{cm}}\] respectively.
Note:
The relation between energy and wavelength is as follows: ${\text{E = }}\dfrac{{{\text{hc}}}}{\lambda}$ . Wavelength and energy are inversely proportional. In hydrogen atoms the lowest energy level has principal quantum number one. The highest energy level can be infinite. The Lyman series lies in the ultraviolet region of the electromagnetic spectrum.
Formula used:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Complete step by step answer:
The formula to determine the wavelength of transitions in hydrogen spectrum is as follows:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Where,
\[\lambda\] is the wavelength.
\[{{\text{R}}_{\text{H}}}\] is the Rydberg constant.
The value of Rydberg constant is $109678\,{\text{c}}{{\text{m}}^{ - 1}}$ .
\[{{\text{n}}_{{\text{lower}}}}\] is the principle quantum of the energy level in which the electron comes.
\[{{\text{n}}_{{\text{higher}}}}\] is the principle quantum of the energy level from which the electron comes.
Lyman series arise when electrons from higher energy levels come to the energy level of principle quantum number one. So, the \[{{\text{n}}_{{\text{higher}}}}\] is \[{{\text{n}}_1}\].
So, the wavelength formula for Lyman series is as follows:
\[\dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]\]
Shortest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the highest energy level. The highest energy level is \[{{\text{n}}_\infty }\].
So, the shortest wavelengths is,
Substitute $109678\,{\text{c}}{{\text{m}}^{ - 1}}$ for Rydberg constant, $n_1=1$ and \[{{\text{n}}_\infty }\] for \[{{\text{n}}_{{\text{higher}}}}\].
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_\infty ^2}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\]
\[\Rightarrow \lambda = \dfrac{{\text{1}}}{{\,109678\,{\text{c}}{{\text{m}}^{ - 1}}}}\]
\[\Rightarrow \lambda = 9.11 \times \,{10^{ - 6}}\,{\text{cm}}\]
So, the shortest wavelength is \[9.11 \times \,{10^{ - 6}}\,{\text{cm}}\].
Longest wavelengths in the hydrogen spectrum of Lyman series means the electron is coming from the next higher energy level. The next higher energy level is second so, the \[{{\text{n}}_{{\text{higher}}}}\] is \[2\].
So, the longest wavelengths is,
Substitute \[{\text{c}}{{\text{m}}^{ - 1}}\] for Rydberg constant and \[{{\text{n}}_2}\] for \[{{\text{n}}_{{\text{higher}}}}\].
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_1^2}} - \dfrac{{\text{1}}}{{{\text{n}}_2^2}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}}\left[ {\dfrac{{\text{1}}}{{{1^2}}} - \dfrac{{\text{1}}}{{{2^2}}}} \right]\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,109678\,{\text{c}}{{\text{m}}^{ - 1}} \times \dfrac{3}{4}\]
\[\Rightarrow \dfrac{{\text{1}}}{\lambda} = \,82258.5\,{\text{c}}{{\text{m}}^{ - 1}}\]
\[\Rightarrow \lambda = \dfrac{{\text{1}}}{{82258.5\,{\text{c}}{{\text{m}}^{ - 1}}}}\]
\[\Rightarrow \lambda = 1.22 \times {10^{ - 5}}\,{\text{cm}}\]
So, the longest wavelength is \[1.22 \times {10^{ - 5}}\,{\text{cm}}\].
Therefore, the shortest and longest wavelengths in hydrogen spectrum of Lyman series are \[9.11 \times \,{10^{ - 6}}\,{\text{cm}}\]and \[1.22 \times {10^{ - 5}}\,{\text{cm}}\] respectively.
Note:
The relation between energy and wavelength is as follows: ${\text{E = }}\dfrac{{{\text{hc}}}}{\lambda}$ . Wavelength and energy are inversely proportional. In hydrogen atoms the lowest energy level has principal quantum number one. The highest energy level can be infinite. The Lyman series lies in the ultraviolet region of the electromagnetic spectrum.
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