In hydrogen atom, an electron jumps from 3rd orbit to the 2nd orbit. calculate the wavelength of the radiation emitted( h=6.63x10-34 j sec

In hydrogen atom, an electron jumps from 3rd orbit to the 2nd orbit. c

1.	In hydrogen atom, an electron jumps from 3rd orbit to the 2nd orbit. calculate the wavelength of the radiation emitted( h=6.63x10-34 j sec
Answer» - ATOMUC STRUCTURECONCEPT :- In Bohr's stationary ORBITS, energy is quantized and the energy if the electron at nth orbit is GIVEN by -13.6 n²/z eV.SOLUTIONSEnergy of electron in 3RD orbit = -13.6 × 1/9Energy of electron in 2nd orbit = -13.6 × 1/4Hence. energy of PHOTON emitted = 13.6 ( 1/4 - 1/9) = 13.6 × 5/36 eV= 1.87 eV.Hence, wavelength of photon emitted = 12400/1.87 Å= 6631 Å