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)

$9 : 4 : 1$

Answer 2

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, $\lambda_p = \frac{9}{R_H} , \lambda_B = \frac{4}{R_H} , \lambda_L = \frac{1}{R_H}$ $\therefore \:\: \lambda_P : \lambda_B : \lambda_L = 9 : 4 : 1$