91Ó°ÊÓ

Consider the Paschen series is the set of line in the hydrogen infrared spectrum.

Consider the formula for the charge to mass ratio in the electron and the proton as follows:

\[\begin{array}{c}\frac{{{q_e}}}{{{m_e}}} = - 1.76 \times {10^{11}}\;\frac{{\rm{C}}}{{{\rm{kg}}}}\\\frac{{{q_p}}}{{{m_p}}} = 9.57 \times {10^7}\;\frac{{\rm{C}}}{{{\rm{kg}}}}\end{array}\]

Here,\[{m_e} = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\]is the mass of the electron and\[{m_p} = 1.67 \times {10^{ - 27}}\;{\rm{kg}}\]is the mass of the proton.

Consider the formula for the Bohr's theory of hydrogen atom.

\[\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right)\]

Here, wavelength of the emitted electromagnetic radiation is \[\lambda \], and the Rydberg constant is \[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\].

Consider the given data:

\[\begin{array}{l}{n_i} = \infty \\{n_f} = 2\\R = 1.097 \times {10^7}\;{m^{ - 1}}\end{array}\]

Thus, wavelength is calculated as

\[\begin{array}{c}\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{{{3^2}}} - \frac{1}{{\;{{\rm{m}}^2}}}} \right)\\\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{9} - 0} \right)\\\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{9}} \right)\\\frac{1}{\lambda } = 1218888.889\;{{\rm{m}}^{ - 1}}\\\lambda = 0.82 \times {10^{ - 6}}\;{\rm{m}}\end{array}\]

In the infrared series, the wavelength mentioned above corresponds to the shortest wavelength.

Hence, it is proved that entire Paschen series is in the infrared part of the spectrum.