Question

The shortest wavelengths of Paschen, Balmer and Lyman series are in the ratio

• A. $9:1:4$
• B. $1:4:9$
• C. $9:4:1$
• D. $1:9:4$

Answer 1

Answer: (C)

The shortest wavelength of Paschen $({\lambda }_P)$ Balmer $({\lambda }_B)$ and Lyman $({\lambda }_L)$ series are given by
$\frac{1}{{\lambda }_P}=\frac{R_H}{3^2};\frac{1}{{\lambda }_B}=\frac{R_H}{2^2}$ and $\frac{1}{{\lambda }_L}=\frac{R_H}{1^2}$
So, ${\lambda }_p=\frac{9}{R_H},{\lambda }_B=\frac{4}{R_H},{\lambda }_L=\frac{1}{R_H}$
$\therefore {\lambda }_P:{\lambda }_B:{\lambda }_L=9:4:1$