Q 12-12-111JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
Taking the wavelength of first Balmer line in hydrogen spectrum ($n = 3$ to $n = 2$) as $660\ \text{nm}$, the wavelength of the $2^{\text{nd}}$ Balmer line ($n = 4$ to $n = 2$) will be
Answer: (B) $488.9\ \text{nm}$
$$\frac{1}{\lambda_1} = R\left(\frac14 - \frac19\right) = \frac{5R}{36},\qquad \frac1{\lambda_2} = R\left(\frac14 - \frac1{16}\right) = \frac{3R}{16}$$
$$\lambda_2 = \lambda_1\times\frac{5/36}{3/16} = 660\times\frac{80}{108} \approx 488.9\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics