Problem 205 - Introductory Chemistry Atoms First

Question

Of the following transitions in the Bohr hydrogen atom, which transition results in the emission of photons with the shortest wavelength? (a) \(n=1 \rightarrow n=6\) (b) \(n=4 \rightarrow n=1\) (c) \(n=9 \rightarrow n=5\) (d) \(n=3 \rightarrow n=6\) (e) \(n=5 \rightarrow n=9\)

Step-by-Step Solution

Answer

The transition with the largest energy difference and therefore the shortest wavelength is (b) \(n=4 \rightarrow n=1\).

Step 1: Recall the energy levels of the Bohr hydrogen atom

The energy levels of the Bohr hydrogen atom can be calculated using the formula: \[E_n = -\frac{13.6 \, eV}{n^2}\] where \(E_n\) represents the energy of the electron in the nth energy level, n is the principal quantum number (any positive integer), and 13.6 eV is the ionization energy for hydrogen.

Step 2: Calculate the energy difference between the transitions

We will now calculate the energy difference, \(\Delta E\), for each transition using the formula: \[\Delta E = E_{final} - E_{initial}\] where \(E_{final}\) is the energy level of the final transition and \(E_{initial}\) is the energy level of the initial transition. (a) \(n = 1 \rightarrow n = 6\): \[\Delta E_a = E_6 - E_1 = -\frac{13.6}{6^2} - (-\frac{13.6}{1^2})\] (b) \(n = 4 \rightarrow n = 1\): \[\Delta E_b = E_1 - E_4 = -\frac{13.6}{1^2} - (-\frac{13.6}{4^2})\] (c) \(n = 9 \rightarrow n = 5\): \[\Delta E_c = E_5 - E_9 = -\frac{13.6}{5^2} - (-\frac{13.6}{9^2})\] (d) \(n = 3 \rightarrow n = 6\): \[\Delta E_d = E_6 - E_3 = -\frac{13.6}{6^2} - (-\frac{13.6}{3^2})\] (e) \(n = 5 \rightarrow n = 9\): \[\Delta E_e = E_9 - E_5 = -\frac{13.6}{9^2} - (-\frac{13.6}{5^2})\]

Step 3: Determine which transition has the largest energy difference

Now we need to find which transition has the largest energy difference. We can either calculate the numerical values for each energy difference and compare them, or we can observe that for the same transition, the largest energy difference occurs when the initial energy level is lower and the final energy level is higher. Comparing the transitions, we can see that (b) n = 4 to n = 1 has the largest energy difference, with the smallest initial energy level and the largest difference between the energy levels.

Step 4: Conclusion

The transition with the largest energy difference and therefore the shortest wavelength is (b) \(n=4 \rightarrow n=1\).

Key Concepts

Energy Levels in the Bohr Hydrogen Atom

The Bohr model of the hydrogen atom presents a valuable picture of how electrons inhabit discrete energy levels. According to this model, an electron orbiting the nucleus can only have specific energy values, which are negative because the electron is in a bound state with the nucleus, indicating that energy needs to be supplied to free the electron.

The energy levels are quantized and can be calculated using the formula:
\[E_n = -\frac{13.6 \, eV}{n^2}\]
where \(E_n\) signifies the electron's energy at the nth level, and \(n\) is the principal quantum number. This number can be any positive integer, and its value indicates the size of the orbit, where a larger \(n\) corresponds to orbits further from the nucleus and with higher potential energy.

In homework problems or textbook exercises involving the Bohr model, you might be asked to calculate the energy difference during electron transitions. This is pivotal in understanding how electrons jump from one energy level to another and is directly related to the emission of light, or photons, observed in experiments like spectral lines.

Photon Emission and the Bohr Model

When an electron in an atom transitions between energy levels, a photon is emitted or absorbed, depending on the direction of the transition. The Bohr model elegantly explains this process in the case of a hydrogen atom. If an electron jumps from a higher energy level to a lower one (for example, from \(n=4\) to \(n=1\)), it emits a photon.

The energy of this emitted photon corresponds to the energy difference between the two levels, given by:
\[\Delta E = E_{final} - E_{initial}\]
The frequency (\(f\)) of the emitted photon is related to its energy (\(E\)) through Planck's equation:
\[E = hf\]
And from this, we can deduce the wavelength (\(\lambda\)) using the speed of light (\(c\)):
\[c = \lambda f\]
Therefore, the shorter the wavelength of the emitted photon, the greater the energy difference between the two energy levels involved in the transition. The transition within the shortest wavelength will be the one with the largest energy difference, as seen in the step-by-step problem solution provided. This phenomena is key in explaining the discrete spectral lines observed for elements and the foundational concept of quantum mechanics.

Quantum Numbers and Electron Configurations

Quantum numbers are the set of numerical values that describe the unique quantum state of an electron within an atom. There are several types of quantum numbers, but the most fundamental one, relevant to the Bohr model, is the principal quantum number, denoted as \(n\). It defines the energy level of an electron within the atom and is integral to determining the size of the orbit in the Bohr model.

As the principal quantum number increases, the electron's orbit becomes larger and its energy less negative (closer to zero). For instance, an electron in a hydrogen atom with \(n=1\) is in the lowest energy state, also known as the ground state. As \(n\) gets larger, the energy levels come closer together, which implies that transitions between high-energy levels result in photon emissions with less energy and consequently longer wavelengths.

Understanding quantum numbers is essential for solving problems related to the structure of atoms, the energy of electrons, and the spectral lines produced. By applying the principal quantum number, students can predict the allowed energy states of electrons and the possible transitions that can occur, as well as the spectral lines that these transitions will produce.