Series in Hydrogen Spectrum Numerical Class-12 Nootan ISC Physics Solution Ch-26 Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Series in Hydrogen Spectrum Numerical Class-12 Nootan ISC Physics Solution Ch-26 Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-26 | Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom |
| Topics | Numericals on Series in Hydrogen Spectrum |
| Academic Session | 2025-2026 |
Numericals on Series in Hydrogen Spectrum
Class-12 Nootan ISC Physics Solution Ch-26 Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom
Que-26: Calculate the wavelength of the first line of the Balmer series of hydrogen. Indicate the transition on an energy-level diagram. (i) What is this line called? (ii) Does it lie in the visible region of the electro-magnetic spectrum?
Ans-
(ii) This line is called Hα.
(iii) This line is found visible in visible region of electromagnetic spectrum.
Que-27: Calculate the wavelengths of the third and the fifth spectral lines of Paschen series in hydrogen spectrum when E = -(2.18 x 10^-18)/n² joule.
Ans-
Que-28: Calculate the wavelengths of the first lines of Balmer, Lyman and Paschen series.
Ans- Suppose that the first line of Balmer series
First line of Lyman series
First line of Paschen series
Que-29: The wavelength of the first line of the Lyman series is λ. What will be the wavelength of the first line of the Balmer series in terms of λ?
Ans- First line of Lyman series
first line of Balmer series
Que-30: The wavelength of the first line of Balmer series is 6563 Å. Find the wavelength of the first line of the Lyman series.
Ans- We know that first line of lyman series
λ = 4/3R ——(1)
and first line of Balmer series
λ’ = 36/5R ——(2)
Que-31: The wavelength of the first member of the Balmer series is 6563 Å. Find the wavelengths of the second and the last member of the series.
Ans- The wavelength of first member of Balmer series
Second member of Balmer series
Dividing (2) by (1) we have
The last member of Balmer series
Dividing eq (3) by (1) we have
=> 5 x 6563 / 9 = 3646 Å
Que-32: The wavelength of the first line of Balmer series is 6563 Å. Calculate (i) Rydberg’s constant, (ii) ionization potential of hydrogen atom.
Ans- first line of Balmer series
and Ionization energy of hydrogen atom
Que-33: The ionization energy of hydrogen atom is 13.6 eV. Calculate Rydberg’s constant for hydrogen.
Ans- We know that
Que-34: The energy transition in hydrogen atom takes place from energy-level n = 8 to n = 1. Calculate the wavelength of the emitted photon. To which spectral series will this photon belong? To which part of electromagnetic spectrum will this photon belong?
Ans- We know that
Que-35: The wavelength of a yellow line of sodium is 5896 Å. Give its wave number.
Ans- We know that
λ = 5896 Å
Que-36: As per given figurer A, B and C are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of two wavelengths (i.e., λ1/λ2) is 7/4n Find the value of n.
Ans- value of n = 5
— : End of Series in Hydrogen Spectrum Numerical Class-12 Nootan ISC Physics Solution :–
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