• # question_answer Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be : [JEE Main 9-4-2019 Morning] A) 889.2 nm                    B) 642.7 nmC) 488.9 nmD) 388.9 nm

$\frac{1}{660}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}$             ....(1) $\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}$                                ....(2) divide equation (1) with (2) $\frac{\lambda }{660}=\frac{5\times 16}{36\times 3}$ $\lambda =\frac{4400}{9}=488.88=488.9nm$ Option [c]