Question

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=4\) to \(n=2,(b)\) from an orbit of radius \(2.12 \AA\) to one of radius \(8.46 \AA\), (c) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell?

Short answer

For the electronic transitions in hydrogen: (a) from \(n=4\) to \(n=2\), energy is emitted with a \(\Delta E = 2.55 \,\text{eV}\), (b) from an orbit of radius \(2.12 \,\text{Å}\) to one of radius \(8.46 \,\text{Å}\), energy is emitted with a \(\Delta E = 2.55 \,\text{eV}\), and (c) when an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell, energy is absorbed with a \(\Delta E = -1.51 \,\text{eV}\).

Step-by-step solution

Determine the energy change during the transition

For the hydrogen atom, the energy of each energy level is given by the formula: \[E_n = -\frac{13.6\,\text{eV}}{n^2}\] where \(n\) is the principal quantum number. To determine the energy change during a transition, we will calculate the difference in energy between the final and initial states: \[\Delta E = E_{final} - E_{initial}\]

Calculate the energy change for the given transition

For the given transition from \(n=4\) to \(n=2\), we will substitute the values into the energy formula and find the energy change: \[\Delta E = E_2 - E_4\] \[= \left(-\frac{13.6\,\text{eV}}{2^2}\right) - \left(-\frac{13.6\,\text{eV}}{4^2}\right)\] \[= -3.4\,\text{eV} - (-0.85\,\text{eV})\] \[\Delta E = 2.55\,\text{eV}\] Since the energy change is positive (\(\Delta E > 0\)), energy is emitted during the transition. Case (b): Transition from an orbit of radius \(2.12\,\text{Å}\) to one of radius \(8.46\,\text{Å}\)

Determine the initial and final energy levels

We are given the radii of the initial and final orbits (in Ångströms). We can use the formula for the radius of the hydrogen atom energy levels to determine the corresponding principal quantum numbers: \[r_n = n^2 a_0\] where \(n\) is the principal quantum number, \(a_0 = 0.529\,\text{Å}\) is the Bohr radius. We'll solve for \(n_{initial}\) and \(n_{final}\): \[r_{initial} = n_{initial}^2 a_0 \Rightarrow n_{initial} = \sqrt{\frac{r_{initial}}{a_0}}\] \[r_{final} = n_{final}^2 a_0 \Rightarrow n_{final} = \sqrt{\frac{r_{final}}{a_0}}\]

Calculate the principal quantum numbers and energy change

Now we will substitute the given radii into the formulas above: \[n_{initial} = \sqrt{\frac{2.12\,\text{Å}}{0.529\,\text{Å}}} \approx 2\] \[n_{final} = \sqrt{\frac{8.46\,\text{Å}}{0.529\,\text{Å}}} \approx 4\] Now we can calculate the energy change for this transition: \[\Delta E = E_4 - E_2\] \[= \left(-\frac{13.6\,\text{eV}}{4^2}\right) - \left(-\frac{13.6\,\text{eV}}{2^2}\right)\] \[\Delta E = -0.85\,\text{eV} - (-3.4\,\text{eV})\] \[\Delta E = 2.55\,\text{eV}\] Since the energy change is positive (\(\Delta E > 0\)), energy is emitted during the transition. Case (c): An electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell

Determine the initial and final energy levels

The initial state is a proton (\(\mathrm{H}^{+}\)), so it has no electrons and no initial energy level. The final state is an electron in the \(n=3\) shell, so the final energy level is \(n_{final} = 3\).

Calculate the energy change for the transition

We need to find the energy change when an electron is added to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell: \[\Delta E = E_3 - E_{initial}\] Since there is no initial energy level, we can consider \(E_{initial} = 0\): \[\Delta E = E_3 - 0\] \[= -\frac{13.6\,\text{eV}}{3^2}\] \[\Delta E = -1.51\,\text{eV}\] Since the energy change is negative (\(\Delta E < 0\)), energy is absorbed during this transition.

Key concepts

Bohr model

The Bohr model is a pivotal concept in understanding how hydrogen atoms behave, especially regarding electronic transitions. This model was introduced by Niels Bohr in the early 20th century. It describes electrons as moving in orbits around the nucleus, similar to planets orbiting the sun. These orbits are energy levels where electrons can exist without radiating energy.
Bohr postulated that electrons can jump between these orbits or energy levels. He explained that when an electron jumps from a higher energy orbit to a lower one, it emits energy in the form of light (a photon). Conversely, when it absorbs energy, it moves from a lower orbit to a higher one. This simple yet powerful model laid the groundwork for quantum mechanics by explaining atomic spectra and the quantized nature of energy levels.
The Bohr model works well for hydrogen-like atoms. While it has limitations with more complex atoms, the model's fundamental principles are still critical to understanding atomic structure and electronic transitions.

Energy levels

Energy levels in an atom are like steps on a staircase. Electrons can be on any step, but they can't be in between steps. These steps are quantified by electron orbits, where each orbit corresponds to a specific energy level. The principal quantum number, represented as \( n \), indicates these levels. The lowest energy level is \( n=1 \), while higher numbers like \( n=2, 3, 4, \) and so on are progressively higher energy levels.
The energy levels become closer together as \( n \) increases. This spacing pattern is crucial for calculating the energy differences when electrons transition between levels. The equation \( E_n = -\frac{13.6\, \text{eV}}{n^2} \) is used for hydrogen, showing energy becomes less negative with increasing \( n \). Lower energy levels (small \( n \)) require more energy to leave, while higher levels are less tightly bound to the nucleus.
Understanding these levels is essential for predicting whether energy will be emitted or absorbed during transitions. As electrons shift between these levels, the energy change determines the absorption or emission of light.

Quantum numbers

Quantum numbers are sets of numerical values that describe an electron's state in an atom. They are crucial for understanding electron configurations and properties in atomic physics, especially in complex atoms beyond hydrogen.

• Principal Quantum Number (\( n \)): This number indicates the main energy level or shell and its typical values are integers (1, 2, 3,...). It primarily determines the electron's energy and distance from the nucleus.
• Angular Momentum Quantum Number (\( l \)): It describes the shape of the orbital and has values ranging from 0 to \( n-1 \). This number influences the electron's angular momentum.
• Magnetic Quantum Number (\( m_l \)): Specifies the orientation of the orbital in space and values range from \(-l\) to \(+l\).
• Spin Quantum Number (\( m_s \)): Represents the electron's spin direction. Its values are typically +1/2 or -1/2.

For the hydrogen atom transitions discussed earlier, primarily the principal quantum number \( n \) and its changes are of interest, determining which energy levels the electron occupies before and after transitions.

Energy emission and absorption

When an electron in a hydrogen atom transitions between energy levels, it may either emit or absorb energy, explaining many observable phenomena in atomic spectra.
**Energy Emission:** This occurs when an electron falls from a higher energy level to a lower one (e.g., from \( n=4 \) to \( n=2 \)). The difference in energy between these levels is released as a photon. The positive energy change (\( \Delta E > 0 \)) indicates emission, as energy exits the atom, often resulting in visible light or other electromagnetic radiation.**Energy Absorption:** Conversely, absorption happens when an electron moves from a lower energy level to a higher one (e.g., from the ground state up to \( n=3 \)). The atom absorbs energy from an external source, resulting in a negative energy change (\( \Delta E < 0 \)). This process requires energy input, allowing electrons to climb to higher energy levels.Effectively, these processes are the basis of emission and absorption spectra seen in various chemical and physical processes, fundamentally explaining how we observe different wavelengths of light from atoms.