Question

In the Bohr model, which of the following electron transitions in a hydrogen atom results in emission of the longest wavelength of light? $$ \begin{aligned} &n=6 \text { to } n=3 \\ &n=5 \text { to } n=3 \end{aligned} $$

Short answer

The transition from \(n=5\) to \(n=3\) results in the emission of the longest wavelength of light.

Step-by-step solution

Calculation energy difference between \(n=6\) to \(n=3\)

Calculate the energy difference using Bohr's formula for the first transition. The final level \(n_f\) is 3 and the initial level \(n_i\) is 6. \[E_1 = -13.6 \, \mathrm{eV} \cdot (1/3^2 - 1/6^2)\]

Calculation energy difference between \(n=5\) to \(n=3\)

Calculate the energy difference for the second transition. The final level \(n_f\) is 3 and the initial level \(n_i\) is 5. \[E_2 = -13.6 \, \mathrm{eV} \cdot (1/3^2 - 1/5^2)\]

Comparison of energy differences

Compare the energy differences. The transition that emits the longest wavelength of light is the one with the smaller energy difference.

Key concepts

Emission of Light Wavelength

The emission of light occurs when an electron in an atom moves from a higher energy level to a lower one, releasing energy in the form of light. The wavelength of this emitted light is inversely proportional to the energy difference between the two levels. That means the greater the energy difference, the shorter the wavelength. This principle is central to understanding how the Bohr model of the atom explains the light emission of hydrogen and other elements.

In the case of hydrogen, when an electron descends from a higher orbit (with principal quantum number, n, being larger) to a lower orbit, it emits light at a specific wavelength that corresponds to the energy difference between those two orbits. According to the problem on hand, the light of the longest wavelength would correspond to the smallest energy transition.

Energy Difference Calculation

The energy of an electron in a particular orbit of a hydrogen atom can be calculated using the formula: \[E = -13.6 \, \mathrm{eV} \cdot \left(\frac{1}{n^2}\right)\], where \(E\) is the energy in electronvolts (eV), and \(n\) is the principal quantum number of the orbit. The negative sign denotes that the electron is bound to the nucleus;

the absolute value of the energy gets smaller (less negative) as the electron moves to orbits farther from the nucleus. To calculate the energy difference when an electron transitions between two levels, the energy of the final state is subtracted from the initial state: \[\Delta E = E_{initial} - E_{final}\], which can be expressed as \[\Delta E = -13.6 \, \mathrm{eV} \cdot \left(\frac{1}{n^2_{initial}} - \frac{1}{n^2_{final}}\right)\].

Utilizing this equation to compare different transitions, one can determine not only the energy differences but also predict the wavelength of the emitted light using the relationship given by Planck's equation.

Hydrogen Atom Electron Transitions

In the Bohr model, electron transitions within a hydrogen atom are quantized events where the electron jumps from one discrete energy level to another. Such transitions are responsible for the characteristic emission spectrum of hydrogen. As mentioned earlier, the energy of each orbit is specific and quantized, following the formula: \[E_n = -13.6 \, \mathrm{eV} \cdot \left(\frac{1}{n^2}\right)\], with each available orbit corresponding to a different principal quantum number \(n\).

When an electron makes a transition from a higher energy level (with a higher value of \(n\)) to a lower one, the energy difference between the two levels is emitted as a photon of light. This photon's energy is precisely equal to the energy difference calculated, and this energy is what determines the light's wavelength. The greater the drop in energy levels, the more energy the photon carries, resulting in a shorter wavelength of light.