
The ground state energy of a hydrogen atom is $ - 13.6\,eV $ . If an electron makes a transition from an energy level $ - 0.85\,eV $ to $ - 3.6\,eV $ , calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does the wavelength belong $ (R = 1.097 \times {10^7}\,{m^{ - 1}}) $ .
Answer
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Hint : In this solution, we will first determine the energy states corresponding to the energy levels given in the question. We will then use Rydberg's formula to determine the wavelength of light emitted in this transition.
Formula used: In this solution, we will use the following formula:
- Energy of electron in hydrogen atom: $ E = \dfrac{{ - 13.6}}{{{n^2}}} $ where $ n $ is the principal number
- Rydberg’s formula: $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) $ where $ {n_1} $ and $ {n_2} $ are the final and initial principal orbit number.
Complete step by step answer
Let us start by determining the principal number corresponding to the energy levels of $ - 0.85\,eV $ and $ - 3.6\,eV $ . We know that the energy of an electron in an energy state $ n $ is calculated as
$ E = \dfrac{{ - 13.6}}{{{n^2}}} $
So, for $ E = - 0.85\,eV $ , substituting the value of energy in the above equation, we get
$ - 0.85 = \dfrac{{ - 13.6}}{{{n^2}}} $ gives us $ n = 3 $
Similarly, for $ E = - 3.6eV $ ,the energy level will be corresponding to $ n = 2 $
So, for the transition from $ n = 3 $ to $ n = 2 $ state, we will observe a wavelength of light determined by the Rydberg’s formula as:
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) $
Substituting $ (R = 1.097 \times {10^7}\,{m^{ - 1}}) $ and taking the inverse of the above equation, we get
$ \lambda = 6563\,{A^0} $
This wavelength falls in the visible spectrum which has a range of $ 4000\,\,{A^0} $ to $ 7000\,{A^0} $ . Visible light corresponds to all the wavelengths of light that can be seen by the human eye.
Note
We must be careful to not calculate the energy difference of the energy levels provided to us to determine the wavelength of light emitted as emitted by transitions must be calculated using Rydberg’s formula. Rydberg’s formula is only applicable to hydrogen-like atoms which are the atoms that only have one valence electron in their valence shell.
Formula used: In this solution, we will use the following formula:
- Energy of electron in hydrogen atom: $ E = \dfrac{{ - 13.6}}{{{n^2}}} $ where $ n $ is the principal number
- Rydberg’s formula: $ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) $ where $ {n_1} $ and $ {n_2} $ are the final and initial principal orbit number.
Complete step by step answer
Let us start by determining the principal number corresponding to the energy levels of $ - 0.85\,eV $ and $ - 3.6\,eV $ . We know that the energy of an electron in an energy state $ n $ is calculated as
$ E = \dfrac{{ - 13.6}}{{{n^2}}} $
So, for $ E = - 0.85\,eV $ , substituting the value of energy in the above equation, we get
$ - 0.85 = \dfrac{{ - 13.6}}{{{n^2}}} $ gives us $ n = 3 $
Similarly, for $ E = - 3.6eV $ ,the energy level will be corresponding to $ n = 2 $
So, for the transition from $ n = 3 $ to $ n = 2 $ state, we will observe a wavelength of light determined by the Rydberg’s formula as:
$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) $
Substituting $ (R = 1.097 \times {10^7}\,{m^{ - 1}}) $ and taking the inverse of the above equation, we get
$ \lambda = 6563\,{A^0} $
This wavelength falls in the visible spectrum which has a range of $ 4000\,\,{A^0} $ to $ 7000\,{A^0} $ . Visible light corresponds to all the wavelengths of light that can be seen by the human eye.
Note
We must be careful to not calculate the energy difference of the energy levels provided to us to determine the wavelength of light emitted as emitted by transitions must be calculated using Rydberg’s formula. Rydberg’s formula is only applicable to hydrogen-like atoms which are the atoms that only have one valence electron in their valence shell.
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