The wavelength limit present in the Pfund series is (R = 1.097 × 107 m-1)

Question

The wavelength limit present in the Pfund series is (R = 1.097 × 107 m-1) (A)  1572 nm (B)  1898 nm (C)  2278 nm (D)  2535 nm

Tardigrade Admin · Accepted Answer

The wavelength for Pfund series is given by (1/λ) = R[(1/52) -(1/n2)] for series limit, n =∞ ∴ (1/λ) = R [(1/25) -(1/∞2)] = (R/25) ∴ λ = (25/R) = (25/1.097 ×107) = 2278 nm

Q. The wavelength limit present in the Pfund series is
$(R = 1.097 \times 1 0^{7} m^{- 1})$

$1572 nm$

$1898 nm$

$2278 nm$

$2535 nm$

The wavelength for Pfund series is given by
$\frac{1}{λ} = R [\frac{1}{5 ^{2}} - \frac{1}{n ^{2}}]$
for series limit, $n = \infty$
$∴ \frac{1}{λ} = R [\frac{1}{25} - \frac{1}{\infty ^{2}}] = \frac{R}{25}$
$∴ λ = \frac{25}{R}$
$= \frac{25}{1.097 \times 1 0 ^{7}}$
$= 2278 nm$

Q. The wavelength limit present in the Pfund series is (R=1.097×107m−1)

1572nm

1898nm

2278nm

2535nm