The Total solution for NCERT class 6-12
A difference of 2.3 eV separatestwo energy levels in an atom. What is the frequency of radiation emitted whenthe atom makes a transition from the upper level to the lower level?
Separation of two energy levelsin an atom,
E = 2.3 eV
= 2.3 ×1.6 × 10−19
= 3.68 ×10−19 J
Let ν bethe frequency of radiation emitted when the atom transits from the upper levelto the lower level.
We havethe relation for energy as:
E = hv
Where,
Hence, the frequency of theradiation is 5.6 × 1014 Hz.