In the above figure D and E respectively represent

Absorption line of Balmer series and the ionization potential of hydrogen

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Absorption line of Balmer series and the wavelength lesser than lowest of the Lyman series

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Spectral line of Balmer series and the maximum wavelength of Lyman series

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Spectral line of Lyman series and the absorption of greater wavelength of limiting value of Paschen series

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Solution

The correct option is A Absorption line of Balmer series and the ionization potential of hydrogen
D is excitation of electron from $2^{n d}$ orbit corresponding to absorption line in Balmer series and E is the energy released to bring the electron from $\infty$ to ground state i.e. ionisation potential.

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Similar questions

Q. Wavelengths of the first lines of the Lyman series, Paschen series and Balmer series in hydrogen spectrum are denoted by

λ_{L}, λ_{P} a n d λ_{B}

respectively. Arrange these wavelength in increasing order.

If $λ_{1}$ and $λ_{2}$ are respectively the wavelengths of the series limit of Lyman and Balmer series of Hydrogen atom, then the wavelength of the first line of the Lyman series of the H-atom is:

Q. The relation between

λ_{3} :

wavelength of series limit of Lyman series,

λ_{2} :

the wavelength of the series limit of Balmer series &

λ_{1} :

the wavelength of first line of Lyman series is

Q. Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is

Q. The difference in the wavelength of the 1st line of Lyman series and 2nd line of Balmer series in a hydrogen atom is :

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In the above figure D and E respectively represent

The correct option is A Absorption line of Balmer series and the ionization potential of hydrogen D is excitation of electron from 2nd orbit corresponding to absorption line in Balmer series and E is the energy released to bring the electron from ∞ to ground state i.e. ionisation potential.