Question

Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 Å is.

• A. Balmer series
• B. Lyman series
• C. Paschen series
• D. Pfund series (Rydberg constant $R = 1.097 \times 10^{7}$per metre)

Answer 1

Answer: (C)

Paschen series

Answer 2

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$ $\Rightarrow \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} = \frac{1}{R\lambda}$

$= \frac{1}{1.097 \times 10^{7} \times 18752 \times 10^{- 10}} = 0.0486 = \frac{7}{144}.$

But $\frac{1}{3^{2}} - \frac{1}{4^{2}} = \frac{7}{144} \Rightarrow n_{1} = 3$ and n₂ = 4

(Paschen series)