Taking the wavelength of first Balmer line in hydrogen spectrum ( n...

From Rydberg’s equation,
$\dfrac{1}{\lambda} =$R\left(\dfrac{1}{{n}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right)$<br/>$\dfrac{1}{660 \times {10}^{- 9}} = $R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{3}^{2}}\right)$
$\dfrac{1}{\lambda} =$R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{4}^{2}}\right)$<br/>$\dfrac{\lambda}{600 \times {10}^{- 9}} = \dfrac{5}{9 \times 4} \times \dfrac{16 \times 4}{12}$<br/>$\lambda = 488.9nm