Question

For each of the following electronic transitions in the hydrogen atom, calculate the energy, frequency, and wavelength of the associated radiation, and determine whether the radiation is emitted or absorbed during the transition: (a) from \(n=4\) to \(n=1,(b)\) from \(n=5\) to \(n=2,(\mathrm{c})\) from \(n=3\) to \(n=6\). Does any of these transitions emit or absorb visible light?

Short answer

For the transition from \(n=4\) to \(n=1\), the energy is emitted, the frequency is \(2.47 \times 10^{15}\ \mathrm{s}^{-1}\), and the wavelength is \(122\ nm\), which is not within the visible light range. For the transition from \(n=5\) to \(n=2\), the energy is emitted, the frequency is \(7.26 \times 10^{14}\ \mathrm{s}^{-1}\), and the wavelength is \(413\ nm\), which is within the visible light range. For the transition from \(n=3\) to \(n=6\), the energy is absorbed, the frequency is \(3.99 \times 10^{14}\ \mathrm{s}^{-1}\), and the wavelength is \(752\ nm\), which is not within the visible light range. Only the transition from \(n=5\) to \(n=2\) emits visible light.

Step-by-step solution

Calculate the energy difference

Using the energy formula and the given initial and final states of the electron, we calculate the energy difference: \(\Delta E = -13.6 \ eV \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = -13.6\ eV\left(1-\frac{1}{16}\right)\) So, \(\Delta E = -10.2\ eV\), which indicates the energy is emitted since \(\Delta E < 0\).

Calculate the frequency

Now we will use the relationship between energy and frequency to find the frequency of the radiation: \(-10.2\ eV = h \cdot \nu\) \(\nu = \frac{-10.2\ eV}{h}\) Where, \(h = 4.135 \times 10^{-15} \ eV \cdot s\) (Planck's constant). \(\nu = \frac{-10.2\ eV}{4.135\times 10^{-15} \ eV \cdot s} = 2.47 \times 10^{15} \ s^{-1}\)

Calculate the wavelength

Using the relationship between frequency and wavelength, we can find the wavelength: \(\lambda = \frac{c}{\nu}\) Where, \(c = 3 \times 10^8\ m/s\) (speed of light). \(\lambda = \frac{3 \times 10^8\ m/s}{2.47 \times 10^{15} s^{-1}} = 1.22 \times 10^{-7} \ m = 122 \ nm\)

Conclusion for (a)

For the transition from \(n=4\) to \(n=1\), the energy is emitted, the frequency is \(2.47 \times 10^{15}\ \mathrm{s}^{-1}\), and the wavelength is \(122\ nm\). This radiation is not in the visible light range. ##b: Transition from \(n=5\) to \(n=2\)## Repeat the steps (1)-(4) for this transition:

Calculate the energy difference

\(\Delta E = -13.6 \ eV \left(\frac{1}{2^2} - \frac{1}{5^2}\right) = -13.6 \ eV\left(\frac{1}{4}-\frac{1}{25}\right)\) So, \(\Delta E = -3.00\ eV\), which indicates the energy is emitted since \(\Delta E < 0\).

Calculate the frequency

\(\nu = \frac{-3.00\ eV}{4.135\times 10^{-15} \ eV \cdot s} = 7.26 \times 10^{14} \ \mathrm{s}^{-1}\)

Calculate the wavelength

\(\lambda = \frac{3 \times 10^8\ m/s}{7.26 \times 10^{14} \ \mathrm{s}^{-1}} = 4.13 \times 10^{-7} \ m = 413 \ nm\)

Conclusion for (b)

For the transition from \(n=5\) to \(n=2\), the energy is emitted, the frequency is \(7.26 \times 10^{14}\ \mathrm{s}^{-1}\), and the wavelength is \(413\ nm\). This radiation is in the visible light range. ##c: Transition from \(n=3\) to \(n=6\)## Repeat the steps (1)-(4) for this transition:

Calculate the energy difference

\(\Delta E = -13.6 \ eV \left(\frac{1}{6^2} - \frac{1}{3^2}\right) = 13.6 \ eV\left(\frac{1}{9}-\frac{1}{36}\right)\) So, \(\Delta E = 1.65\ eV\), which indicates the energy is absorbed since \(\Delta E > 0\).

Calculate the frequency

\(\nu = \frac{1.65\ eV}{4.135\times 10^{-15} \ eV \cdot s} = 3.99 \times 10^{14}\ \mathrm{s}^{-1}\)

Calculate the wavelength

\(\lambda = \frac{3 \times 10^8\ m/s}{3.99 \times 10^{14}\ \mathrm{s}^{-1}} = 7.52 \times 10^{-7}\ m = 752 \ nm\)

Conclusion for (c)

For the transition from \(n=3\) to \(n=6\), the energy is absorbed, the frequency is \(3.99 \times 10^{14}\ \mathrm{s}^{-1}\), and the wavelength is \(752\ nm\). This radiation is not in the visible light range. In conclusion, only the transition from \(n=5\) to \(n=2\) emits visible light with a wavelength of \(413\ nm\).

Key concepts

Energy Levels of Hydrogen Atom

The hydrogen atom serves as a cornerstone for understanding the electronic structure of atoms. A hydrogen atom consists of a single electron bound to a nucleus with one proton. The quantum mechanical model provides us with a way to calculate the atom's energy levels, which are characteristically quantized. This means that the electron can only reside in certain allowed energy states.

The distribution of these energy levels is represented by the principal quantum number, denoted as 'n.' Each level is associated with a specific energy value, given by the formula \( E_n = -\frac{13.6 eV}{n^2} \). Here, -13.6 eV represents the energy of the electron in the lowest energy state (n=1), also called the ground state.

When the electron transitions between these quantized energy levels, it either absorbs or emits a photon with energy exactly equal to the difference between these levels. This is additionally illustrated in the exercise solution above, where the energy changes for different transitions were calculated.

Quantum Mechanical Model

Understanding the behavior of electrons in atoms requires the quantum mechanical model, which departs significantly from classical physics' predictability. This model incorporates principles like wave-particle duality and quantization, asserting that particles like electrons exhibit both wave-like and particle-like properties.

According to the quantum mechanical model, electrons do not travel in defined orbits but rather in regions of space called 'orbitals,' each with a distinct energy level. These orbitals are defined by four quantum numbers: the principal quantum number (n), angular momentum (l), magnetic (m), and spin (s). In the context of hydrogen's energy levels, we generally focus on the principal quantum number, which defines the major energy levels of the atom.

Spectral Lines Calculation

The discrete energy levels in a hydrogen atom lead directly to the formation of spectral lines. A spectral line is formed when an electron makes a transition between energy levels, which was exemplified in the exercise solution where we calculated the frequency and wavelength of the emitted or absorbed radiation.

The frequency (\(u\)) of the radiation is related to the energy difference (\(\Delta E\)) between the two levels by the equation \(\Delta E = h \cdot u\), where 'h' is the Planck's constant. Furthermore, we can express the wavelength (\(\lambda\)) in terms of the frequency through the equation \(\lambda = \frac{c}{u}\), where 'c' is the speed of light.

As demonstrated in the original exercise, only specific transitions correspond to wavelengths in the visible spectrum, which is key when analyzing the radiation of hydrogen or any other element through spectroscopy. Understanding these concepts is essential for students to grasp the foundational principles of atomic structure and quantum mechanics.