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=$1.03\times {10}^7{m}^{-1}]$

Answer 1

Bohr’s third postulate successfully explains emission lines. It states that ‘Whenever 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 the energy difference between the initial and final states’.

Thus,

$E_1-E_f=hv=\frac{hc}{\lambda }$

The Rydberg formula for the Balmer series is given as

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

‘R’ is a constant called the Rydberg constant and its value is $1.03\times {10}^7{m}^{-1}$.The $H_{\alpha }$-line of the Balmer series is obtained when an electron jumps to the second orbit $\left(n_{f}2\right)$from the third orbit $\left(n_{i}2\right)$

$\frac{1}{\lambda }=1.03\times {10}^7(\frac{1}{2^2}-\frac{1}{3^2})\frac{1}{\lambda }=1.03\times {10}^7(\frac{1}{4}-\frac{1}{9})\frac{1}{\lambda }=1.03\times {10}^7\left(\frac{9-4}{9\times 4}\right)\frac{1}{\lambda }=1.03\times {10}^7\left(\frac{5}{36}\right)\lambda =\frac{36}{5\times 1.03\times {10}^7}\lambda =6.99\times {10}^{-7}\lambda =699\times {10}^{-2}\times {10}^{-7}\lambda =699nm$

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