Question

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2$^{nd}$ Balmer line (n = 4 to n = 2) will be :

• A. $889.2 \,nm$
• B. $642.7 \,nm$
• C. $488.9 \,nm$
• D. $388.9 \,nm$

Answer 1

Answer: (C)

$\frac{1}{660} = R \bigg(\frac{1}{2^2} - \frac{1}{3^2}\bigg) \, = \, \frac{5R}{36} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1)$
$\frac{1}{\lambda} = R \bigg(\frac{1}{2^2} - \frac{1}{4^2}\bigg) \, = \, \frac{3R}{16} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(2)$
divide equation (1) with (2)
$\frac{\lambda}{660} = \frac{5 \times 16}{6 \times 3}$
$\lambda = \frac{4400}{9} = 488.88 = 488.9 nm$
Option (3)