Question

Sate Bohr’s quantisation condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the e.m. spectrum it belongs.

Answer 1

Bohr’s quantization condition of angular momentum: 

According to Bohr, electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total angular momentum of the revolving electron is an integral multiple of \(\frac{h}{2\pi}\)Where h is planck’s constant. 

Thus, the angular momentum of the orbiting electron is quantized. 

As, angular momentum of electron = mvr

∴ for any permitted (stationary) orbit

Where n = 1,2,3…… called as principal quantum number 

Shortest wavelength, n₂= ∞, n₁ = 4

This wavelength belongs to infrared region of e.m. spectrum.