Excitation and lonisation Potentials 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.
Excitation and lonisation Potentials 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 Excitation and lonisation Potentials |
| Academic Session | 2025-2026 |
Numericals on Excitation and lonisation Potentials
Numerical Class-12 Nootan ISC Physics Solution Ch-26 Atom, Origin of Spectra : Bohr’s Theory of Hydrogen Atom
Que-20: The wavelength of the first spectral line of sodium is 5896 Å. What is the first excitation potential of sodium atom?
Ans- Given,
Que-21: The first excitation potential of mercury is 4.86 V. A mercury atom, after absorbing a photon, reaches in its first excited state. Find the wavelength of the absorbed photon.
Ans- E = 4.86 eV = 4.86 x 1.6 x 10^-19 Joule
Suppose that the wavelength is λ then
Que-22: The first and second excitation potentials of hydrogen atom are respectively 10.2 and 12.09 V. Draw, on this basis, the energy-level diagram of the atom and show all possible emission transitions in it and mark their corresponding wavelengths.
Ans-
Que-23: The ionization potential of hydrogen atom is 13.6 V. In the lowest energy-level, this atom is ionized by absorbing a photon of 800 Å. What will be the kinetic energy of the released electron?
Ans- Ionization potential = 13.6 V
=> 13.6 x 1.6 x 10^-19 Joule
λ = 8000 Å = 800 x 10^-10 m
∴ Kinetic energy of released electron => 15.47 – 13.6 = 1.87 eV
Que-24: The ionization-potential of hydrogen atom is 13.6 V. Find out: (i) wavelength of Hβ line of Balmer series, (ii) shortest wavelength of Lyman series.
Ans- Ionization potential = 13.6 eV
for line HB ΔE2 = -0.85 – (-3.4) = 2.55 eV
because for line HB atom n = 4 energy level to n = 2 energy level
Shortest wavelength of Lyman series
Que-25: Calculate the wavelength of the Hα line of hydrogen spectrum. The energy levels of hydrogen atom are given by the expression En = -(13.6/n²) eV.
Ans- For Hα, the atom turns to n = 3 to n = 2 then,
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