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)
S safeer

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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