Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main2013ChemistryStructure of AtomActual

The wave number of the first emission line in the Balmer series of H -Spectrum is : ( R = Rydberg constant ) :

Options

  1. A5 36 R
  2. B9 400 R
  3. C7 6 R
  4. D3 4 R

Correct answer

A. 5 36 R

Step-by-step solution

v = RZ ^2 ( 1 2^2 - 1 3^2 ) = R ( 1 4 - 1 9 )= 5 R 36

Practice Structure of Atom on Quantrex Academy →

More from Structure of Atom

The 2s and the 2p orbital energies of hydrogen atom are E_ 2s ( H ) and E_ 2p ( H ) , respectively. The 2s and the 2p orbital energies of lithium atom are E_ 2s ( Li ) and E_ 2p ( 2026X ^ a+ and Y ^ b+ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers n = 1 and n = 2 of X ^ a+ is . 2026Consider two radiations of wavelengths 1. ₁ = 2000 Å 2. ₂ = 6000 Å The ratio of the energies of these two radiations ( E₁ E₂ ) is ________ (Nearest integer). 2026The Bohr radius of a hydrogen like species is 70.53 pm. The species and the stationary state (n) are respectively (Given : Hydrogen atom Bohr radius is 52.9 pm) 2026If shortest wavelength of hydrogen atom in Lyman series is x , then longest wavelength in Balmer series of He ^+ is: 2026Match the List-I with List-II List-I Orbital List-II Radial nodes and nodal plane A. 2s I. 1 Radial node + two nodal planes B. 3s II. 1 Radial node + one nodal plane C. 3p III. 2 R 2026Which of the following statement(s) is/are true ? A. If two orbitals have the same value of (n + l) , the orbital with lower value of n will have lower energy. B. Energies of the o 2026Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. A. n = 3, l = 2, m = +1 B. n = 4, l = 0, m = 0 C. n = 6, l = 1, m = 0 D. n = 5, l = 1, 2026 Full Structure of Atom list All JEE Main PYQs