Question

The shortest wavelength of Paschen series in hydrogen spectrum is $8182\mathring{A}$. The first member of the Paschen series is nearly

• A. $15400\mathring{A}$
• B. $12200\mathring{A}$
• C. $13400\mathring{A}$
• D. $18700\mathring{A}$
• E. $16700\mathring{A}$

Answer 1

Answer: (D)

Wave length of the spectral series of Paschen series is
$\lambda =\frac{9n^2}{\left(n^2-9\right)R}$
where $n=4,5,6,\dots$
Shortest wavelength of Paschen series ${\lambda }_s=\frac{9}{R}$
Longest wavelength of Paschen series
${\lambda }_L=\frac{16}{7}\times \left(\frac{9}{R}\right)=\frac{16}{7}{\lambda }_s$
$=\frac{16}{7}\times 8182=18700\mathring{A}$