Question

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In hydrogen atom, an electron undergoes transition from 3rd excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong.
(ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases.
(i) These transitions belong to :
1. Balmer series,
2. Lyman series​

Answer 1

Above answer is correct!!

MARK BRAINLIEST answer

Answer 2

Here for wavelength λ1

N1=4 and N2= 3

and for λ2

N1= 3 and N2= 2

We have :-

$\frac{hc}{ \:λ} = - 13.6( \frac{1}{n2} - \frac{1}{n1} ) \\ so \: for \: λ1 \\ \frac{hc}{ \:λ1 } = - 13.6( \frac{1}{ {4}^{2} } - \frac{1}{ {3}^{2} } ) = - 13.6( \frac{7}{144} )$

Similarly for λ2

$\frac{hc}{ \:λ2} = - 13.6( \frac{5}{36} )$

Hence , putting both the equations we get :-

$\frac{λ1}{ \:λ2} = \frac{20}{7}$

Here for wavelength λ1

N1= 4 and N2= 3

And for λ2

N1=3 and N2= 2

Hence the ratio is 20:7 .

These transitions belongs to Lyman series.

❥ Learn More:-

The spectral line for the Lyman series is the transitions from n >= 2 to the n = 1 state, for Balmer series its from n >= 3 to the n = 2 state and the Paschen series are transitions from n >= 4 to n = 3 state. ... The radiation emitted in such a transition corresponds to the spectral line in the atomic spectra of H-atom.

Lyman and Balmer series are hydrogen spectral line series that arise from hydrogen emission spectra. The key difference between Lyman and Balmer series is that Lyman series forms when an excited electron reaches the n=1 energy level whereas Balmer series forms when an excited electron reaches the n=2 energy level.