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 Hα line. [Given: Rydberg constant R = 1.03 × 107 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 × 107 m−1.
The Hα-line of the Balmer series is obtained when an electron jumps to the second orbit (nf = 2) from the third orbit (ni = 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.