The wave number of hydrogen atom in Lyman series is $82200c{{m}^{-1}}$.The electron goes from
(A) ${{n}_{3}}$to ${{n}_{2}}$
(B) ${{n}_{2}}$ to ${{n}_{1}}$
(C) ${{n}_{4}}$ to ${{n}_{3}}$
(D) none of these

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Hint: The energy emitted by the photon during transition is directly proportional to the wavenumber. During transition, when energy is emitted,${{n}_{1}}$will be less than ${{n}_{2}}$, that is transition will be from higher orbital to lower orbital.

Complete answer:
 A hydrogen atom consists of an electron orbiting its nucleus. The electromagnetic force between the electron and proton leads to a set of quantum states for the electron, each having its own energy.Spectral emissions occur when an electron jumps, or there is transition of electron from a higher state to a lower state. To distinguish the two states,the higher state is commonly designated as n and the lower state is commonly designated as n’. Energy is emitted when the electron is transitioned from higher state to lower state and energy is absorbed when the electron goes from lower to higher state. The energy emitted corresponds to the energy difference between the two states. Since the energy of the state is fixed, the transition will always produce a photon with the same energy.
The Rydberg formula was given to calculate the energy differences between the level in the bohr model, and hence the wavelengths of emitted/absorbed photons, the Rydberg formula is given as
follows,$\dfrac{1}{\lambda }={{Z}^{2}}R\left( \frac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)$
Where, R is the Rydberg constant = $109678c{{m}^{-1}}$
Z is the atomic number, taken 1 for hydrogen atom.
${{n}_{1}}$= lower energy state (1 for Lyman series)
${{n}_{2}}$= higher energy state${{n}_{2}}$

Wavelength is reciprocal of wavenumber, therefore, $\dfrac{1}{\lambda }=82200c{{m}^{-1}}$,
Putting the values in the equation, we get,
     \[82200=109678\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\]
     \[\dfrac{82200}{109678}=\left( \dfrac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)\]
     \[\dfrac{3}{4}=\left( \dfrac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)\]
     \[\dfrac{1}{{{n}_{2}}^{2}}=1-\dfrac{3}{4}=\frac{1}{4}\]
     \[\dfrac{1}{{{n}_{2}}^{2}}=1-\dfrac{3}{4}=\frac{1}{4}\]
     \[{{n}_{2}}=2\]
 Therefore, the electron jumps from the second orbital (${{n}_{2}}$) to ground state (${{n}_{1}}$).
So, the correct answer is “Option B”.

Note: The Lyman series includes the lines due to transitions from an outer orbit n > 1 to the orbit n' = 1.The Balmer series includes the lines due to transitions from an outer orbit n > 2 to the orbit n' = 2.