Question

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by\[\frac{{\bf{1}}}{{\bf{\lambda }}}{\bf{ = R}}\left( {\frac{{\bf{1}}}{{{\bf{n}}_{\bf{f}}^{\bf{2}}}}{\bf{ - }}\frac{{\bf{1}}}{{{\bf{n}}_{\bf{i}}^{\bf{2}}}}} \right){\bf{?}}\]It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

Short answer

The average percentage difference found is 0.025 %. The phenomenon is amazing as it shows the error between the actual value and the wavelength of the calculated value of the energy level is also approximately equal to zero.

Step-by-step solution

Define the formulas:

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

Define the average percentage difference found between these wavelength numbers and those predicted by\[\frac{{\bf{1}}}{{\bf{\lambda }}}{\bf{ = R}}\left( {\frac{{\bf{1}}}{{{\bf{n}}_{\bf{f}}^{\bf{2}}}}{\bf{ - }}\frac{{\bf{1}}}{{{\bf{n}}_{\bf{i}}^{\bf{2}}}}} \right)\]

For Balmer series,\[{n_f} = 2\]

Case 1: For\[{n_i} = 3\]

Substitute\[R = 1.097 \times {10^7}{m^{ - 1}},{n_f} = 2\]and\[{n_i} = 3\]

The Wavelength is calculated as:

\[\begin{array}{c}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{9}} \right)\\\frac{1}{\lambda } = 1.524 \times {10^6}\;{{\rm{m}}^{ - 1}}\\\lambda = 656.53\;{\rm{nm}}\end{array}\]

Also the known value is656.5nm

Thus, percentage difference is calculated as:

\[\frac{{656.5\;{\rm{nm}} - 656.3\;{\rm{nm}}}}{{656.5\;{\rm{nm}}}} \times 100 = 0.030\% \]

Case 2: For\[{n_i} = 4\]

Substitute\[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}},\;{n_f} = 2,\;{n_i} = 4\].

The Wavelength is calculated as:

\[\begin{array}{c}\frac{1}{\lambda } = \left( {1.097 \times {{10}^{7\;}}{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{{16}}} \right)\\\frac{1}{\lambda } = 2.057 \times {10^6}\;{{\rm{m}}^{ - 1}}\\\lambda = 486.17\;\;{\rm{nm}}\end{array}\]

Also, the known value is486.3nm.

Thus, percentage difference is determine as:

\[\frac{{486.3\;{\rm{nm}} - 486.14\;{\rm{nm}}}}{{486.3\;{\rm{nm}}}} \times 100 = 0.032\% \]

Case 3: For\[{n_i} = 5\]

Substitute\[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}},\;{n_f} = 2,\;{n_i} = 5\]

The Wavelength is calculated as

\[\begin{array}{c}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{{25}}} \right)\\\lambda = 434\;{\rm{nm}}\end{array}\]

Also, the known value is434.2nm.

Thus, percentage differenceis calculated as:

\[\frac{{434.2\;{\rm{nm}} - 434\;{\rm{nm}}}}{{434.2\;{\rm{nm}}}} \times 100 = 0.046\% \]

Case 4: For\[{n_i} = 6\]

Substitute\[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}},\;{n_f} = 2,\;{n_i} = 6\]

The Wavelength is calculated as

\[\begin{array}{c}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{6^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{{36}}} \right)\\\lambda = 410.2\;{\rm{nm}}\end{array}\]

Also, the known value is 410.3

Thus, percentage differenceis determined as:

\[\frac{{410.3\;{\rm{nm}} - 410.2\;{\rm{nm}}}}{{410.3\;{\rm{nm}}}} \times 100 = 0.024\% \]

Hence, the average percentage difference found is 0.025 %. The phenomenon is amazing as it shows the error between the actual value and the wavelength of the calculated value of the energy level is also approximately equal to zero.