Question

Do the Balmer and Lyman series overlap? To answer this, calculate the shortest-wavelength Balmer line and the longest-wavelength Lyman line.

Short answer

No, Balmer and Lyman series does not overlap.

Step-by-step solution

What is difference between balmer and lyman series? 

The Lyman series occurs for the excited electron at n=1 energy level and the Blamer series occurs for the excited electron at n=2 energy level.

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

Determine whether the Balmer and Lyman series overlap or not. 

Solve forbalmer series,

Consider the given data.

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

Thus, wavelength is calculated as

\[\begin{array}{l}\frac{1}{{{\lambda _B}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{{{2^2}}} - \frac{1}{{\;{{\rm{s}}^2}}}} \right)\\\frac{1}{{{\lambda _B}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{4} - 0} \right)\\\frac{1}{{{\lambda _B}}} = 2742500\;{{\rm{m}}^{ - 1}}\\{\lambda _B} = 3.6463 \times {10^{ - 7}}\;{\rm{m}}\end{array}\]

Solve further as:

\[{\lambda _B} = 364.63\;{\rm{nm}}\]

Solve for the Lyman series

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

Substitute the values and solve as:

\[\begin{array}{c}\frac{1}{{{\lambda _L}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)\\\frac{1}{{{\lambda _L}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{1}{1} - \frac{1}{4}} \right)\\\frac{1}{{{\lambda _L}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{{4 - 1}}{4}} \right)\\\frac{1}{{{\lambda _L}}} = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\left( {\frac{3}{4}} \right)\end{array}\]

Solve further as:

\[\begin{array}{c}{\lambda _L} = 1.215 \times {10^{ - 7}}\;{\rm{m}}\\{\lambda _L} = 121.5 \times {10^{ - 9}}\;{\rm{m}}\\{\lambda _L} = 122\;\;{\rm{nm}}\end{array}\]

Thus, \[{\lambda _L} \ne {\lambda _B}\]

Therefore, Balmer and Lyman series does not overlap.