Question
State Bohr’s quantisation condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.
✓
Answer
According to Bohr's quantisation, the electrons revolve around the nucleus only in those orbits for which the angular momentum is the integral multiple of
$\frac{\text{h}}{2\pi}$
$\text{L}=\frac{\text{nh}}{2\pi}$
For Bracket series $\text{n}_2=\infty,$
$\frac{1}{\lambda}=\text{R}_\text{H}\text{Z}^2\Big\{\frac{1}{4^2}-\frac{1}{\infty}\Big\}$
$\frac{1}{\lambda}=\frac{\text{R}_\text{H}}{16}$
$\lambda=\frac{16}{\text{R}_\text{H}}=14.58\times10^{-7}\text{m}$
This wavelength belongs to the infra-red region.
$\frac{\text{h}}{2\pi}$
$\text{L}=\frac{\text{nh}}{2\pi}$
For Bracket series $\text{n}_2=\infty,$
$\frac{1}{\lambda}=\text{R}_\text{H}\text{Z}^2\Big\{\frac{1}{4^2}-\frac{1}{\infty}\Big\}$
$\frac{1}{\lambda}=\frac{\text{R}_\text{H}}{16}$
$\lambda=\frac{16}{\text{R}_\text{H}}=14.58\times10^{-7}\text{m}$
This wavelength belongs to the infra-red region.
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