Question

The first member of the paschen series in hydrogen spectrum of wave length is $18800\stackrel{\text{o}}{A}$ The short Wavelength limit of Paschen series is:

• A. $1215\stackrel{\text{o}}{A}$
• B. $6560\stackrel{\text{o}}{A}$
• C. $8225\stackrel{\text{o}}{A}$
• D. $12850\stackrel{\text{o}}{A}$

Answer 1

Answer: (C)

The wavelength of radiation emitted in, Paschen series $\frac{1}{\lambda }=R\left[\frac{1}{3^2}-\frac{1}{n^2}\right]$ For first member, n = 4 $\therefore$ $\frac{1}{18800}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$ $\therefore$ $\frac{1}{18800}=R\times \left(\frac{16-9}{144}\right)$ or $R=\frac{144}{18800\times 7}$ ?(i) For shortest wavelength limit (series limit) $n=\infty$ $\frac{1}{\lambda }=R\left(\frac{1}{3^2}-\frac{1}{{\infty }^2}\right)$ $=\frac{144}{18800\times 7}\times \frac{1}{9}$ [from Eq. (i)] Hence, $\lambda =\frac{18800\times 7\times 9}{144}=8225\stackrel{\text{o}}{A}$