Question

The wavelength limit present in the Pfund series is

$(R = 1.097 \times 10^{7}\text{m }\text{s}^{- 1})$

• A. 1572 nm
• B. 1898 nm
• C. 2278 nm
• D. 2535 nm

Answer 1

Answer: (C)

2278 nm

Answer 2

The wavelength for Pfund series is given by

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{5^{2}} - \frac{1}{n^{2}} \right\rbrack$

For series limit $n = \infty$

$\therefore\frac{1}{\lambda} = R\left\lbrack \frac{1}{25} - \frac{1}{\infty^{2}} \right\rbrack = \frac{R}{25}$

$\therefore\lambda = \frac{25}{R} = \frac{25}{1.097 \times 10^{7}} = 2278nm.$