Home
Class 12
PHYSICS
A difference of 2.3 eV separates two ene...

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Promotional Banner

Similar Questions

Explore conceptually related problems

The ionization energy of H atom is 13.6 eV. what is the energy of Li2+ ion in first excited state.

What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from an energy level with n = 6 to an energy level with n = 3 ?

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes trnsition from an energy level with n = 4 to an energy level with n = 2 ?

The energy levels of an atom are as shown in Fig.1.23 Which transition corresponds to emission of radiation of maximum wavelength? .

How do the energy gaps between successive electron energy levels in an atom very from low to high n values ?

Given below is some famous number associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which belongs. 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen, known as Lamb shift).

A hydrogen like species (atomic number Z) is present in a higher excited state of quantum number n. This excited atom can make a transitionn to the first excited state by successive emission of two photons of energies 10.20 eV and 17.0 eV respectively. Altetnatively, the atom from the same excited state can make a transition to the second excited state by successive of two photons of energy 4.25 eV and 5.95 eVv respectively. Determine the value of Z.