Question

calculate the maximum and minimum wavelength for paschen series of hydrogen atom.( R=1.097x10^7 m^-1)

Answer 1

hey mate your ans refer to attachment

Answer 2

Given:

R = 1.097 × 10⁷ m⁻¹

To find:

Calculate the maximum and minimum wavelength for the Paschen series of the hydrogen atom.

Solution:

$\frac{1}{\lambda} = R[\frac{1}{n^2} - \frac{1}{m^2} ]$

The maximum wavelength for the Paschen series of the Hydrogen atom

n = 3, m = 4

$\frac{1}{\lambda _ m_a_x} = R[\frac{1}{3^2} - \frac{1}{4^2} ] \\\\\frac{1}{\lambda _ m_a_x} = 1.097 * 10^7 [\frac{1}{9} - \frac{1}{16} ]\\\\\frac{1}{\lambda _ m_a_x} = 1.097 * 10^7 (\frac{16 - 9}{16 * 9} )\\\\\frac{1}{\lambda _ m_a_x} = 1.097 * 10^7 (\frac{7}{144} )\\\\\lambda _ m_a_x = \frac{144}{1.097 * 10^7 * 7}$

λmax = 18.75 × 10⁻⁷ m

λmax = 1875 nm

The minimum wavelength for the Paschen series of the Hydrogen atom

n = 3, m = ∞

$\frac{1}{\lambda _ m_i_n} = R[\frac{1}{3^2} - \frac{1}{\infty ^2} ] \\\\\frac{1}{\lambda _ m_i_n} = 1.097 * 10^7 [\frac{1}{9}]\\\\\frac{1}{\lambda _ m_i_n} = \frac{1.097 * 10^7}{9}\\\\ \lambda _ m_i_n = \frac{9}{1.097 * 10^7}$

λmin = 8.20 × 10⁻⁷ m

λmin = 820 nm