Question

Calculate the shortest and longest wavelength of transition in Balmer series of atomic hydrogen.

Answer 1

for Balmer series, the lowest orbit will be 2.

Shortest $n_2=\infty$
$\frac{1}{\lambda }=R_H(\frac{1}{2^2}-\frac{1}{{\infty }^2})1^2=\frac{R_H}{4}$
$\lambda =\frac{4}{R_H}$
longest $n_2=3$
$\frac{1}{\lambda }=R_H(\frac{1}{2^2}-\frac{1}{3^2})1^2$
$\lambda =\frac{36}{5R_H}$