Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202616 April 2026Morning ShiftChemistryStructure of AtomActual

What is the energy of an electron in a hydrogen atom in a stationary state corresponding to n = 2 ?

Options

  1. A-5.45 10⁻¹⁹ J
  2. B-2.40 10⁻¹⁹ J
  3. C-4.35 10⁻¹⁸ J
  4. D-6.70 10⁻¹⁸ J

Correct answer

A. -5.45 10⁻¹⁹ J

Step-by-step solution

The energy of an electron in the n th stationary state of a hydrogen atom is given by the expression: E_n = -R_H ( 1 n^2 ) where R_H is the Rydberg constant for energy, which is equal to 2.18 10⁻¹⁸ J. For n = 2 : E₂ = -2.18 10⁻¹⁸ ( 1 2^2 ) J E₂ = - 2.18 10⁻¹⁸ 4 J E₂ = -0.545 10⁻¹⁸ J E₂ = -5.45 10⁻¹⁹ J Answer: -5.45 10⁻¹⁹ J

Practice Structure of Atom on Quantrex Academy →

More from Structure of Atom

The frequency of photon which is emitted during a transition of electron of He ^+ ion from fifth energy level to third energy level will be: 2026Which of the following statement is correct ? 2026The wavelength of a particular electron transition for He ⁺ is 100 nm. The wavelength (in Å ) of H atom for the same transition is 2025The energy of second Bohr orbit of hydrogen atom is -3.4 eV. The energy of the fourth Bohr orbit of the He ⁺ ion will be 2025The work function of Cu is 7.68 10⁻¹⁹ ~J . If photons of wavelength 221 nm are made to strike the surface of the metal, the kinetic energy (in J) of the ejected electrons will be ( 2025In an element with atomic number (Z) 25 , the number of electrons with ( n+l ) value equal to 3 and 4 are x and y respectively. The value of (x+y) is 2025a , b , c , d are electromagnetic radiations. Frequencies of a , b are 3 10¹⁵ ~Hz , 2 10¹⁴ ~Hz , respectively, whereas wavelength of c, d are 400 ~nm , 750 ~nm , respectively. The 2025The number of electrons with magnetic quantum number, m _l=0 in the elements with atomic numbers Z=24 and Z=29 are respectively 2025 Full Structure of Atom list All MHT CET PYQs