i. Using Bohr's second postulate of quantisation of orbital angul… - Vidyadip

Question

i. Using Bohr's second postulate of quantisation of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n-times the de-Broglie wavelength associated with it.
ii. The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

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Answer

i. Only those orbits are stable for which the angular momentum of revolving electron is an integral multiple of $\left(\frac{h}{2 \pi}\right)$ where $h$ is the planck's constant.
According to Bohr's second postulate
$\begin{array}{l}m v r_n=n \frac{h}{2 \pi} \Rightarrow 2 \pi r_n=\frac{n h}{m v} \\ \text { But } \frac{h}{m v}=\frac{h}{p}=\lambda \text { (By de Broglie hypothesis) } \\ \therefore 2 \pi r_n=n \lambda\end{array}$
ii. For third excited state, n = 4
For ground state, n = 1
Hence possible transitions are
$\begin{array}{l} n _{ i }=4 \text { to } n _{ f }=3,2,1 \\ n _{ i }=3 \text { to } n _{ f }=2,1 \\ n _{ i }=2 \text { to } n _{ f }=1\end{array}$

Total number of transitions = 6
$\begin{array}{l} E _C- E _{ R }=\frac{h c}{\lambda_1} \ldots( i ) \\ E _{ B }- E _{ A }=\frac{h c}{\lambda_2} \ldots \text { (ii) } \\ E _{ C }- E _{ A }=\frac{h c}{\lambda_a} \ldots \text { (iii) }\end{array}$
Adding (i) and (ii), we have
$E _{ C }- E _{ A }=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \ldots$ (iv)
From (iii) and (iv), we have
$\begin{array}{l}\frac{h c}{\lambda_3}=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \Rightarrow \frac{1}{\lambda_3}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \\ \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\end{array}$

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