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
$\lambda=\frac{9 n^{2}}{\left(n^{2}-9\right) R}$
where $n=4,5,6, \ldots$
Shortest wavelength of Paschen series $\lambda_{s}=\frac{9}{R}$
Longest wavelength of Paschen series
$\lambda_{L}=\frac{16}{7} \times\left(\frac{9}{R}\right)=\frac{16}{7} \lambda_{s}$
$=\frac{16}{7} \times 8182=18700 \, \mathring{A}$