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To what series does the spectral lines of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series 486.1 and 410.2 nm?

$(a)\;2^{nd}  to  3^{rd} level\qquad(b)\;4^{th}  to  2^{nd} level\qquad(c)\;5^{th}  to  1^{st} level\qquad(d)\;6^{th}  to  4^{th} level$

1 Answer

Given
$\lambda_1 = 486.1\times10^{-9}$m = 486.1$\times10^{-7}$ cm
$\lambda_2 = 410.2\times10^{-9}$m = 410.2$\times10^{-7}$ cm
$\overline V = \overline V_2 - \overline V_1$ = $\large\frac{1}{\lambda_2}$ - $\large\frac{1}{\lambda_1}$
$=R_H[\large\frac{1}{2^2}$ -$\large\frac{1}{n_2^2}-R_H[\large\frac{1}{2^2} -\large\frac{1}{n_1^2}]$
$\overline V = R_H[\large\frac{1}{n_1^2} -\large\frac{1}{n_2^2}]$
For I case of Balmer Series :
$\large\frac{1}{\lambda_1}= R_H[\large\frac{1}{2^2} -\large\frac{1}{n_1^2}]$
= 109678 [$\large\frac{1}{2^2} - \large\frac{1}{n_1^2} ]$
Or [$\large\frac{1}{486.1\times10^{-7}}= 109678 [\large\frac{1}{2^2}-\large\frac{1}{n_1^2}]$
For II case of Balmer series:
$\large\frac{1}{\lambda_2}= R_H[\large\frac{1}{2^2} -\large\frac{1}{n_2^2}]$
= 109678 [$\large\frac{1}{2^2} - \large\frac{1}{n_2^2} ]$
Or [$\large\frac{1}{410.2\times10^{-7}}= 109678 [\large\frac{1}{2^2}-\large\frac{1}{n_2^2}]$
$\therefore n_2$ = 6
$\therefore$ given transition occurs from $6^{th}$ level to $4^{th}$ level
Hence answer is (d)
answered Jan 28, 2014 by sharmaaparna1
 

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