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

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)

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