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**Hint:**The excitation and de-excitation of atoms into various orbits involves absorption and emission of some source of energy. \[\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\]

In emission spectra de-excitation occurs, and loss of energy occurs as the photons are emitted.

**Complete step by step solution:**

In the question it is given that, while we give a light of wavelength $\lambda $to a excited state of atom, re-excitation occurs and then when the light source is removed de-excitation occurs which yields the spectral lines and there are 10 spectral lines in the spectra with 7 of wavelength greater than the incident light and 2 of them with lesser wavelength of incident wavelength.

Now let’s give a theoretical approach to find the answer, before that we should note that in the question it is mentioned that emission spectrum is obtained to get an emission spectra the prerequisite is that the atom should be in an excited state.

And now let’s see the options given, the orbits given,

For option (A) if the atoms are in the third orbit i.e. n=3, then by giving energy its excitation and de-excitation takes place. As the atom is in the third excited orbit, the atom will be unstable. So to obtain stability, it de-excites back to ground state and the transitions occurs between n=3 to n=2 and n=2 to n=1, so there will be two de-excitations which will result in two spectral lines as it emission takes place from the excited orbit, energy loss through radiations like photons happens and the energy will be less than the incident wavelength.

For option (B) ,the atoms are in the fourth orbit and emissions involved are ,n=4 to n=3,n=3 to n=2,n=2 to n=1 ,so there are three de-excitations taking place to return to the ground state and there will be three spectral lines in the emission spectra.

For option I C), the atoms are in n=5 state and there are four emissions involving to reach the ground state which, n=5 to n=4, n4 to n=3, n=3 to n=2, n=2 to n=1

For option (D), the atoms are present in the de-excited state with the n value 5 i.e. in the fifth orbit and the transitions which results in the spectrum spectra with less wavelength than the incident wavelength are n=6 to n=5, n=5 to n=4, n4 to n=3, n=3 to n=2, n=2 to n=1

**So the correct option for this question is option (A)**

**Note:**For doing the problems related to the wavelength and spectral lines and transitions of hydrogen atom we can use the Rydberg formula, which is

\[\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)\], where ${{R}_{H}}$ IS Rydberg’s constant with value 109677$c{{m}^{-1}}$

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