Question

The first member of the Balmer series of hydrogen atom has a wavelength of $6561\mathring{A}$. The wavelength of the second member of the Balmer series (in nm) is ____________.
Given $15122$

Answer 1

Answer: 486.00

For Balmer series,
$\frac{1}{\lambda }=R_H\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)$
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}$
$\frac{{\lambda }_2}{6561}=\frac{5/36}{3/16}$
${\lambda }_2=4860\mathring{A}$
$=486nm$