(solved)Question 2.17 of NCERT Class XI Chemistry Chapter 2

wavenumber $\displaystyle \overline{\nu} = \frac{1}{\lambda}$

longest wavelength ⇒ shortest wavenumber

but wavenumber for balmer series H atom is

$\displaystyle \overline{\nu} = 109677 \bigg(\frac{1}{2^2} - \frac{1}{n^2}\bigg) \text{cm}^{-1}$

where n = 3, 4, 5, and so on

$\displaystyle \implies \overline{\nu}$ is shortest if $\displaystyle n = \text{ lowest }= 3$

$\displaystyle \therefore \overline{\nu} = 109677 \bigg(\frac{1}{2^2} - \frac{1}{3^2}\bigg) \text{cm}^{-1}$

$\displaystyle = 109677 \times \frac{5}{36} \text{ cm}^{-1} $

$\displaystyle = 15233 \text{ cm}^{-1} = 15233 \times 10^2 \text{ m}^{-1}$

$\displaystyle = 1.5233 \times 10^6 \text{ m}^{-1} \:\underline{Ans}$