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_{α} line. [Given: Rydberg constant R = 1.03 × 10^{7} m^{−1}]

#### Solution

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_t=hv=(hc)/lambda`

The Rydberg formula for the Balmer series is given as

`1/lambda=R(1/n_t^2-1/n_1^2)`

‘R’ is a constant called the Rydberg constant and its value is 1.03 × 10^{7} m^{−1}.

The H_{α}-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).

`1/lambda=1.03xx10^7(1/2^2-1/3^3)`

`1/lambda=1.03xx10^7(1/4-1/9)`

`1/lambda=1.03xx10^7((9-4)/(9xx4))`

`1/lambda=1.03xx10^7(5/36)`

`lambda=36/(5xx1.03xx10^7)`

λ = 6.99 x 10^{-7}

λ = 699 x 10^{-2 }x 10^{-7}

λ = 699 nm

The value of λ lies in the visible region.