91Ó°ÊÓ

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 λ , and the Rydberg constant is \[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\].

Consider the given values:

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

Substitute the values and solve as:

\[\begin{array}{c}\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - {\rm{1}}}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)\\\frac{1}{\lambda } = 8.2275 \times {10^6}\;{{\rm{m}}^{ - 1}}\\\lambda = 1.215 \times {10^{ - 7}}\;{\rm{m}}\end{array}\]

Wavelength is 122 nm.

The wavelength of the layman series is\[1.215 \times {10^{ - 7}}\;{\rm{m}}\].