Home
Class 12
PHYSICS
Consider a hydrogen-like atom whose ener...

Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_(n) = - (13.6 Z^(2))/(n^(2))`
When this excited makes a transition from excited state to ground state , most energetic photons have energy
`E_(max) = 52.224 eV`. and least energetic photons have energy
`E_(min) = 1.224 eV`
Find the atomic number of atom and the initial state or excitation.

Promotional Banner

Topper's Solved these Questions

  • ATOMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION H (Multiple True-False)|2 Videos
  • ATOMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION I (Subjective)|4 Videos
  • ATOMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION F (Matrix-Match)|4 Videos
  • ALTERNATING CURRENT

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section-J) (Aakash Chailengers Questions)|2 Videos
  • COMMUNICATION SYSTEMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION D (Assertion-Reason)|10 Videos

Similar Questions

Explore conceptually related problems

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom /ion

Energy E of a hydrogen atom with principle quantum number n is given by E = (-13.6)/(n^(2)) eV . The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately

Energy E of a hydrogen atom with principle quantum number n is given by E = (-13.6)/(n^(2)) eV . The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately

The electron in a hydrogen atom at rest makes a transition from n = 2 energy state to the n = 1 ground state. find the energy (eV) of the emitted photon.

In a hydrogen atom, if energy of an electron in ground state is - 13.6 eV , then that in the 2^(nd) excited state is :

The energy of an atom or ion in the first excited state is -13.6 eV. It may be

In a hydrogen atom , If the energy of electron in the ground state is -x eV ., then that in the 2^(nd) excited state of He^(+) is

If the energy in the first excited state in hydrogen atom is 23.8 eV then the potential energy of a hydrogen atom in the ground state can be assumed to be

In hydrogen atom, energy of first excited state is - 3.4 eV . Then, KE of the same orbit of hydrogen atom is.

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength lambda . If R is the Rydberg constant, then the principal quatum number n of the excited state is