Question

Find the wavelength of the third line in the Lyman series, and identify the type of EM radiation.

Short answer

The wavelength of the third line in the Lyman series is 97.234 nm and it corresponds to the wavelength of UV radiation.

Step-by-step solution

Determine the formulas:

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 \[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\].

Find the wavelength of the third line in the Lyman series

Substitute the values and solve for the third line in the layman series as:

\[\begin{array}{l}\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{4^2}}}} \right)\\\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{1} - \frac{1}{{16}}} \right)\\\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{{16 - 1}}{{16}}} \right)\\\frac{1}{\lambda } = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{{15}}{{16}}} \right)\end{array}\]

Solve further as:

\[\begin{array}{l}\lambda = 9.7234 \times {10^{ - 8}}\;{\rm{m}}\\\lambda = 97.234 \times {10^{ - 9}}\;{\rm{m}}\\\lambda = 97.234\;\;{\rm{nm}}\end{array}\]

Wavelength is 97.234 nm

The above wavelength obtained lies in UV radiation.

Therefore, the required wavelength of the third line in the Lyman series is 97.234 nm and it corresponds to the wavelength of UV radiation.