Question

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 1

TOPIC :- ATOMUC STRUCTURE

CONCEPT :- 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.

SOLUTIONS

Energy of electron in 3rd orbit = -13.6 × 1/9

Energy of electron in 2nd orbit = -13.6 × 1/4

Hence. 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