Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202431 Jan 2024Morning ShiftPhysicsAtomic PhysicsActual

If the wavelength of the first member of Lyman series of hydrogen is λ . The wavelength of the second member will be

Options

  1. A27 32 λ
  2. B32 27 λ
  3. C27 5 λ
  4. D5 27 λ

Correct answer

A. 27 32 λ

Step-by-step solution

For Lyman series: For first member, 1 λ = 13 . 6 Z 2 h c 1 1 2 - 1 2 2 ….. (i) For second member, 1 λ ' = 13 . 6 Z 2 h c 1 1 2 - 1 3 2 ….. (ii) On dividing (i) and (ii), we get λ ' = 27 32 λ

Practice Atomic Physics on Quantrex Academy →

More from Atomic Physics

Consider a hydrogen atom with v_k , r_k , and K_k denoting the velocity, orbital radius and kinetic energy of the electron in the k^ th orbit, respectively. The electron undergoes 2026The ratio of momentum of the photons of the 1^ st and 2^ nd line of Balmer series of Hydrogen atoms is / . The possible values of and are:- 2026In the hydrogen atom, the electron makes a transition from the higher orbit ( i ) to a lower orbit ( f ). The ratio of the radius of the orbits in given by r_i : r_f = 16 : 4 . The 2026In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because A. The size of gold nucleus is very small as compared to the size of gold atom 2026Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2^ nd and 4^ th orbits of hydrogen atom _______. 2026Angular momentum of an electron in a hydrogen atom is 3h , then the energy of the electron is _______ eV. 2026Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 10⁵ ~m / s and 2.5 10⁵ ~m / s respectively. If the radii of these orbits are nearly same then the possib 2026In hydrogen atom spectrum, (R Rydberg's constant ) A. the maximum wavelength of the radiation of Lyman series is 4 3 R B. the Balmer series lies in the visible region of the spectr 2026 Full Atomic Physics list All JEE Main PYQs