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

For Balmer series,
$\frac{1}{\lambda}=R_{H}\left(\frac{1}{2^{2}}-\frac{1}{n^{2}_{2}}\right)$
$\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)}$
$\frac{\lambda_{2}}{6561}=\frac{5 / 36}{3 / 16}$
$\lambda_{2}=4860\,\mathring{A}$
$=486\,nm$