State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong.

 
 
 

Answers (1)

According to the Bohr's quantization condition:- The angular momentum of an electron in an orbit around the hydrogen atom has to be an integral multiple of plank's constant divided by 2 \pi

Mathematically it can be written as:-

L=\frac{nh}{2\pi }

For the Brackett series, the wavelength can be given by the formula :

\frac{1}{\lambda }=R\left ( \frac{1}{4^{2}}-\frac{1}{n^{2}} \right )

'n' has to be maximum for the wavelength to be shortest, that is n=\infty

Hence, 

\frac{1}{\lambda }=1.097\times 10^{7}m^{-1}\left ( \frac{1}{16}-0 \right )

\lambda =1.46\times 10^{-6}m

This wavelength (\lambda =1.46\times 10^{-6}m) belongs to the infrared part of the electromagnetic spectrum

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