Question

A wavelength of 4.653 µm is observed in a hydrogen spectrum for a transition that ends in then_f= 5 level. What was ni for the initial level of the electron?

Short answer

The initial energy level of the transition is found to be 7.

Step-by-step solution

Define the Rydberg constant

The Rydberg constant is the lowest-energy photon capable of ionising the hydrogen atom from its ground state, or the highest wavenumber of any photon that may be released from the hydrogen atom.

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 the ni  for the transition occurred. 

Consider the wavelength and energy level equation is solved as:

\[\begin{array}{l}\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right)\\\frac{1}{\lambda } = R\left( {\frac{1}{{25}} - \frac{1}{{n_i^2}}} \right)\\\frac{1}{{n_i^2}} = \frac{1}{{25}} - \frac{1}{{\lambda R}}\end{array}\]

Substitute the values and solve as:

\[\begin{array}{l}\frac{1}{{n_i^2}} = \frac{1}{{25}} - \frac{1}{{1.097 \times {{10}^7}\;{m^{ - 1}} \times 4.653 \times {{10}^{ - 6}}\;m.}}\\{n_i} \approx 7\end{array}\]

Therefore, the initial energy level of the transition is found to be 7 .