Question

State Bohr's postulate of hydrogen atom which successfully explains the emission lines in the spectrum of hydrogen atom.
Use Rydberg formula to determine the wavelength of $H_{\alpha }$ line.
[Given: Rydberg constant $R=1.03{10}^7{m}^{-1}$]

Answer 1

Bohr's third postulate successfully explains emission lines. It states that "Whenever ever an electron jumps from one of its specified non-radiating orbit to another such orbit, it emits or absorbs a photon whose energy is equal to energy difference between the initial and final states".

Thus,

$E_i-E_f=hv=\frac{hc}{\lambda }$

The Rydberg formula for the Balmer series is given as,

$\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

${}^{\prime }{R}^{\prime }$ is a constant called the Rydberg constant and its value is $1.03$ X

${10}^7{m}^{-1}$.

The $H_{\alpha }$ line of the Balmer series is obtained when an electron jumps to the second orbit ($n_f=2$) from the third orbit ($n_i=3$).

$\frac{1}{\lambda }=1.03$ X ${10}^7(\frac{1}{2^2}-\frac{1}{3^2})$

$\frac{1}{\lambda }=1.03$ X ${10}^7(\frac{1}{4}-\frac{1}{9})$

$\frac{1}{\lambda }=1.03$ X ${10}^7\left(\frac{9-4}{9\ast 4}\right)$

$\frac{1}{\lambda }=1.03$ X ${10}^7\left(\frac{5}{36}\right)$

$\lambda =6.99$ X ${10}^{-}7$

$\lambda =699$ X ${10}^{-2}$ X ${10}^{-7}$

$\lambda =699nm$

The value of $\lambda$ lies in the visible region.